8. THE D- AND f-BLOCK ELEMENTS – LONG ANSWER TYPE QUESTIONS

8. THE D- AND f-BLOCK ELEMENTS

Q. 1. What is lanthanoid contraction ? What are the consequences of lanthanoid contraction ?

Ans⇒ A group of fourteen elements following lanthanum i.e., from ₅₆Ce to ₇₁ Lu placed in 6th period of long form of periodic table is known as lanthanoids (or lanthanide series). These fourteen elements are represented by common general symbols ‘Ln’. In these elements, the last electron enters the 4f-subshells (pre pen ultimate shell). It may be noted that atoms of these elements have electronic configuration with 6s² common but with variable occupancy of 4f level. However, the electronic configuration of all the tripositive ions (the most stable oxidation state of all lanthanoids) are of the form 4f1 (n = 1 to 14 with increasing atomic number.) These elements constitute one of the two series to inner transition elements of f-block.

Lanthanoid contraction : In the lanthanoide series with the increase in atomic number, atomic radii and ionic radii decrease from one element to the other, but this decrease is very small. The regular small decrease in atomic radii and ionic radii of lanthanides with increasing atomic number along the series is called lanthanoid contraction.

Cause of lanthanoid contraction : When one move from ₅₈ Ce to ₇₁ Lu along the lanthanide series nuclear charge goes on increasing by one unit every time. Simultaneously an electron is also added which enters to the inner f subshell. The shielding effect of f-orbitals in very poor due to their diffused shape. It results in the stronger force of nuclear attraction of the 4f electrons and the outer electrons causing decrease in size.

Consequences of lanthanoid contraction : (i) Similarly in the properties of second and third transition series e.g., Sr and Hf; Nb and Ta; Mo and W. This resemblance is due to the similarity in size due to the presence of lanthanoids in between.

(ii) Similarity among lanthanoids : Due to the very small change in size, all the lanthanoids resemble one another in chemical properties.

(iii) Decrease in basicity : With the decrease in ionic radii, covalent character of their hydroxide goes in increasing from Ce(OH)₃ to Lu(OH₃ and so base strength goes on decreasing.

Q. 2. Explain giving reasons :

(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst.

Ans⇒ (i) Becausē transition metal atom or ions have unpaired d-electrons in their configuration.

(ii) Because atoms is these elements are held together by strong metallic bonds as both ns electrons and (n − 1)d electrons take part in the metallic bonding.

(iii) Because most transition metal ions contain one or more unpaired electrons in their d-orbitals, (i.e., d-d transition are possible).

(iv) Because of their ability of adopt multiple oxidation states and to form complexes. V₂O₅ (in contact process for manufacture of H₂SO₄) finely divided iron (in Haeber’s process for NH₃ manufacture) and Ni (in catalytic hydrogenation) are examples of their good catalytic activities.

Q.  3. How is the variability in oxidation states of transition metals different from that of the non transition metals ? Illustrate with example.

Ans⇒ They variability in oxidation states is a fundamental characteristic of transition elements and it arises due to incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity. For example, vandium, V show the oxidation states of +2, +3, +4 and +5. Similarly Cr shows oxidation state of +2, +3, +4, +5 and +6; Mn shows all oxidation states from +2 to +7.

This is contrasted with variability of oxidation states of non-transition elements where oxidation states generally differ by units of two. For example, S shows oxidation states of -2, +2, +6 while P shows +3 and +5 oxidation states. Halogenes like Cl, Br and I show oxidation states of -1, +1, +3, +5 and +7 states. In non transition elements variability of oxidation states is caused due to unpairing of electrons in ns or np orbitals and their promotion to np or nd vacant orbitals.

Q.  4. Describe the preparation of potassium dichromate from iron chromite one. What is the effect of increasing pH on a solution of potassium dischromate ?

Ans⇒ following steps are involved in preparation of K₃Cr₂O₇ from iron chromite (FeCr₂O₄) ore :

(i) preparation of sodium chromate : The chromate ore (FeO.Cr₂O₃ ) is finely powdered and mixed with sodium carbonate and quick lime and then heated to redness in the reverbetary

3FeO Cr₂O₃ + O₂ → 2Fe₂O₃ + 4Cr₂O₃
[4Na₂CO₃ + 2Cr₂O₃ + 3O₂ → 4Na₂CrO₄ + 4CO₂] × 2
4FeO.Cr₂O₃ + 8Na₂CO₃ + 7O₂ → 4Na₂CrO₄ + 2Fe₂O₃ + 😯₂]

The mass is then extracted with water, when sodium chromate is completely dissolved while Fe₂O₃ is left behind.

(ii) Conversion of sodium chromate into sodium dichromate (NaCr₂O₇) : The sodium chromate extreme with water in previous steps is actified.

3Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O

On cooling Na₂SO₄ separates out as Na₂SO₄. 10 H₂O and Na₂Cr₂O₇ is remains in solution.

(iii) Conversion of Na₂Cr₂O₇ into K₂Cr₂O₇ : The solution containing Na₂Cr₂O₇ is treated with KCl.

Na₂Cr₂O₇ + KCI → K₂Cr₂O₇ + 2NaCl

Sodium chloride (NaCl) being less soluble separafes out on cooling. On crystalising the remaining solution, Orange coloured crystals of K₂Cr₂O₇ separate out.

Effect of Change of pH : When pH of solution of K₂Cr₂O₇ is increased slowly the medium changes from acidic to basic. The chromates and dichromates are interconvertible in aqueous solution depending spon pH of solution.

Cr₂O + H₂O 2CrO+ 2H⁺

dichromate ion at low pH chromate ion at high pH (orange in acidic medium) (yellow in alkaline medium)
At low pH (acidic medium), K₂Cr₂O₇ solution is oranged coloured while at higher pH (alkaline medium) it changes to yellow due to formation of chromate ions.

Q. 5. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with :

(a) Iodide (b) Iron solution and (c) H2S.

Ans⇒ Potassium dichromate, K₂Cr₂O₇ is a strong oxidising and is used as primary standard in volumetric analysis involving oxidation of iodides, ferrous ion and S2-ions etc. In acidic solution its oxidising action can be represented as follows :

Cr2O + 14H⁺ + 6e → 2Cr³⁺ + 7H₂O; (E⁺ = 1.33V)

(a) It oxidises potassium iodide to iodine.

Cr2O + 14H⁺61^– → 2Cr³ + 7H₂O + 3l₂

(b) It oxidises iron (II) salt to iron (III) salt

Cr2O + 14H⁺ + 6Fe²⁺ + 2Cr³⁺ = 6Fe³⁺ + 7H₂O

(c) It oxidises H₂SO to S

Cr₂O + 8H⁺ + 3H₂S → 2Cr³⁺ + 7H₂O + 3S

Q. 6. Describe the preparation of potassium permanganate. How does the acidified permanganate solution reacts with (a) iron (II) ions (b) SO₂ and (c) oxalic acid ? Write the ionic equations for the reactions.

Ans⇒ Preparation of KMnO₄ from pyrolusite ore (MnO₂) involves the following steps :

(i) Fusion of ore with alkali. in presence of air : Pyrolusite ore is fused with alkali in the presence of air when potassium mangnate is obtained as green mass.

2MnO₂ + 4KOH + O₂ + 2K₂MnO₄ + 2H₂O
.                               (green mass)

The green mass is dissolved in water to obtain aqueous solution of potassium manganate. The insoluble impurities of sand and other metal oxides are removed by filtration.

(ii) Oxidation of mangnate into permanganate : The aqueous solution of K₂MnO₄ is oxidised
lectrolytically or by using ozone or Cl₂ to obtain rotassium permanganate. The process is carried out till Green colour disappear and solution acquires distinet pink colour.

MnO₄^2- → MnO₄^– + e^– (oxidation at anode) (at anode) pink green colour

H₂O + e^– → 1/2H₂ + (OH)^– (reduction at cathode) (at cathode)

or, 2K₂MnO₄ + Cl₂ → 2KMnO₄ + 2KCl
(green colour)               (pink colour)

Potassium permanganate is crystalised out from the solution.

Oxidising Properties : It acts as a powerful oxidising agent in different media differently. In acidic medium, it oxideses iron (II) salt to iron (II) salt, SO₂ to H₂SO₄ and oxalic acid to CO₂ and H₂O.

(a) It oxidises iron (II) salt to iron (III) salts.

2MnO₄^2- + 16H⁺ + 10Fe²⁺ → 2Mn²⁺ 8H₂O + 10Fe³⁺

(b) It oxidised sulphur dioxide to sulphuric acid.

2MnO₄^– + 5SO₂ + 2H₂O → 5SO₄^2- + 2Mn²⁺ 4H⁺

(c) It oxidises oxalic acid to CO₂ and H₂O 2mnO₄^– + 16H + ⁺ 5C₂O₄² → 2Mn²⁺ + 8H₂O + 10CO₂

Q. 7. Use Hund’s rule to derive the electronic configuration of Ce³⁺ and calculate its magnetic moment on the basic of ‘spin only’ formula.

Ans⇒ The electronic configuration of Ce and Ce³⁺ ions is :

Ce(Z = 58) = ₅₄ [Xe]4f¹ 5d¹ 6s²
Ce³⁺ = ₅₄ [Xe]4f ¹

The no. of unpaired electron = 1
‘Spin only formula for magnetic moment of a species,

∴    Magnetic moment of ce³⁺

B.M = √3B.M. = 1.732 B.M

Q. 8. Compare the chemistry of the actinoids with that of lanthanoids with reference to (i) Electronic configuration (ii) Oxidation states and (iii) Chemical reactivity.

Ans⇒ (i) Electronic configuration : All the actinoids are believed to have the electron configuration of 7s² and variable occupancy of the 5f and 5d subshell. The fourteen elements are formally added to 5f.
Similarly all the lanthanoids are believed to have the electron fonfiguration of 6s2 and variable occupancy of 4f level.

(ii) Oxidation state : There is greater range of oxidation states in actinoids which is attributed to the fact that the 5f, 6d and 7s levels are having comparable energies and take part in giving different oxidation states.

Whereas in lanthanoids + 3 oxidation state is predominant because of the 6s² and only one f-orbital taking part. However, occasionally +2 and +4 ions in solution or in solid compounds are also obtained :

(iii) Chemical reactivity : The actinoids are highly reactive metals like lanthanoids especially when finely divided. Both actinoids and lanthanoids react with boiling water to give a mixture of oxide. and hydride and combination with non-metals takes place at moderate temperature.

Q. 9. Describe briefly the following physical chemical properties of transition metals :

(i) Ionic radii.
(ii) Magnetic behaviour.
(iii) Formation of interstitial compounds.

Ans⇒ (i) Ionic radii : For ions carrying identical charges the ionic radii decreases slowly with increase in the atomic numbers across a given series of the transition elements. The decrease in the size is due to the increase nuclear charge across the series while the electrons being added into the d-orbitals of the penultimate shell.

(ii) Magnetic properties : Most of the compounds of transition elements are paramagnetic in nature and are attracted by the magnetic field.

The transition elements involve the partial filling of d-subshells. Most of the transition metal ions or their compounds have unpaired electrons in d-subshell and therefore they give rise to paramagnetic character. The larger the number of unpaired electrons in a substance, the greater in the paramagnetic character.

(iii) Interstitial compounds : Transition metals form interstitial compounds with lighter elements such as hydrogen, boron, carbon and nitrogen. In these compounds small atoms of these elements occupy the interstitial spaces of the metal lattices. These are non stoichiometric compound and cannot be given any definite formulae. They have essentially the same chemical properties as the parent metals but differ in physical properties such as density, and hardness.

Q. 10. What 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 ?

Ans⇒ The d-block elements are called TRANSITION ELEMENTS, because these elements represent change (or transition) in properties from most positive electropositive s-block elements to least electropositive p-block elements. They are sandwitched between s-block of the periodic table (highly reactive metals) and p-block (mostly non-metals). The last electron enters penultimated ((n – 1) d) subshells in the atoms of these elements. These elements have partially filled d-subshells in their elementtary states or in their commonly occurring oxidation states. Zn, Cd, Hg do not possess partially filled inner. d-orbitals either in their elementarty states or in their common oxidation states.

Zn = 30 = [Ar] 3d¹⁰ 4s²                         Zn (II) = 3d¹⁰
Cd = 48 = [Kr] 4d¹⁰ 5s²                        Cd (II) = 4d¹⁰
Hg = 80 = [Xe] 4ƒ¹⁴5d¹⁰6s²                Hg (II) = 5 d¹⁰

∴    Zn, Cd, Hg (belonging to 12th group of the periodic table) may not be regarded as transition elements. However, they have been treated along with other transition elements because they show similarities in some of their properties like complex formation etc. with them.

Characteristics :

1. They are hard and brittle metals.

2. They have high melting and boiling points and have higher heats of vaporisation than non-transition elements.

3. These elements have very high densities as compared to I and II group metals.

4. The first IEs of d-block elements are higher than those of s-block elements, but are lesser than those of p-block elements.

5. They are electropositive in nature.

6. Most of them with electronic configurations of (n – 1) d^1-9 form coloured compounds. Some of them with (n-1) d⁰ or (n – 1) d¹⁰ form white compounds.

7. Because of their small size and large charge density, these ions take part in a large number of complex formations.

8. Because of the tendency of inner d-orbital electrons to take part in compound formations along with ns^1-2 electrons, they exhibit variable oxidation states.

9. Because of the presence of one or more unpaired electrons, their compounds are generally paramagnetic.

10. Because of similar sizes of their atoms, these elements form alloys e.g. Brass is an alloy of Cu and Zn

11. Because of the availability of empty d-orbitals, these metals form Interstitial compounds with small sized atoms like H, B, C, N etc.

12. Most of the transition metals such as Mn, Ni, Co, Cr, V, Pt etc. and their compounds particularly oxides like V₂O₅, Cr₂O₃, COoO, NiO, KMnO₄, K₂ Cr₂O₇ are used as catalysts. They use their inner d-orbitals to form bonds with the reactants.

Q. 11. Compare the chemistry of actinoids with that of the lanthanoids with special reference to :

(a) electronic configuration,
(b) atomic and ionic sizes,
(c) oxidation state and
(d) chemical reactivity.

Ans⇒ (a) Eletronic configuration of actinoids and lanthanoids : Both are Inner-Transition Elements. The inner transition elements or f-block elements are those which have partly filled f-subshells of the antipenultimate orbit in their elementary or ionic state. In both lanthanoids and actinoids the last electron enters (n-2) ƒ subshell

Lanthanoids have general electronic configuration of [Xe] 4ƒ^0-14 5d^0-1 6s²
where [Xe] stands for Xe core.

Actinoids have the general electronic configuration of (Rn] 5ƒ^0-14 6d^0-2 7 S² where (Rn) stands for radon core.

Last electrons enters 4ƒ in the case of lanthanoids and it enters 5f in the case of actinoids.

(b) Atomic sizes of Ln vary from 187 pm for La to .173 for Yb. Their ionic sizes [Ln³⁺] vary from 106 pm for La³⁺ to 85 pm for Lu³⁺, where as the size of actinoids vary from 111 pm for Ac³⁺ to 98 pm for Cf³⁺ and from 99 pm for Th⁴⁺ to 86 pm for Cf⁴⁺.

(c) Oxidation states : Whereas Lanthanoids show a dominant oxidation state of + 3 [+2, +4, are exceptions] actinodes show variable oxidations state of +2, +3, +4, +5, +6, +7.

(d) The chemical reactivity etc. are summarised below :

Comparision of Lanthanoide and Actinoide Series : Elements of lanthanoide and actinoides resemble in many respects but they do differ in some respects as shown below :

Class 12th Chemistry Long Type Question