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.
MnO42- → 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.
2MnO42- + 16H+ + 10Fe2+ → 2Mn2+ 8H2O + 10Fe3+
(b) It oxidised sulphur dioxide to sulphuric acid.
2MnO4– + 5SO2 + 2H2O → 5SO42- + 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.
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) d1-9 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 ns1-2 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] 4ƒ0-14 5d0-1 6s2
where [Xe] stands for Xe core.
Actinoids have the general electronic configuration of (Rn] 5ƒ0-14 6d0-2 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 :
|1||Mainly exhibit + 3 oxidation state.||1||They also mainly exhibit + 3 oxidation state.|
|2||They exhibit lantha noide contraction.||2||They exhibit actinoide contraction.|
|3||They show ion- exchange behaviour.||3||They also show ion exchange behaviour.|
|1||In addition to + 3 oxidation state, they exhibit + 2 and + 4 oxidation states only.||1||In addition to + 3 oxidation state, they show +4, + 5, + 6 and +7 oxidation states.|
|2||Most of their ions are colourless.||2||Most of their ions are coloured.|
|3||They do not form complexes easily.||3||They have much greater tendency to form complexes.|
|4||They do not form oxocations.
||4||They form oxocations such as UO22+, PuO22+ and UO+.|
|5||Their compounds are less basic.
||5||Their compounds are more basic.|
|6||Except promethium, they are non- radioactive.
||6||They are radioactive.|
|7||Their magnetic properties can be easily explained.
||7||Their magnetic properties cannot be easily explained.|
Class 12th Chemistry Long Type Question
|S.N||CHEMISTRY LONG TYPE QUESTION 2022|
|6||GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS|
|7||THE P-BLOCK ELEMENTS|
|8||THE D- AND f-BLOCK ELEMENTS|
|10||HALOALKANES AND HALOARENES|
|11||ALCOHOLS, PHENOLS AND ETHERS|
|12||ALDEHYDES, KETONES AND CARBOXYLIC ACIDS|
|16||CHEMISTRY IN EVERYDAY LIFE|
|17||DISTINGUISH BETWEEN PAIR|