Bihar Board - Class 12 Chemistry - Chapter 8: The d-and f-Block Elements Long Answer Question

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

Ans: A group of fourteen elements following lanthanum i.e., from ₅₆Ce to ₇₁Lu  placed in 6th the period of long form of the 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 penultimate 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 lanthanides) 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 lanthanide 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 lanthanide contraction : When one move from ₅₆Ce to ₇₁Lu  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 are very poor due to their diffused shape. It results in the stronger force of nuclear attraction of the 4f electrons and the outer electrons cause decrease in size.
Consequences of lanthanide 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 lanthanides in between.
(ii) Similarity among lanthanides : Due to the very small change in size, all the lanthanides 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(OH)₃ and so base strength goes on decreasing.

Q. 2. Explain giving reasons :
(i) Transition metals and many of their compounds show paramagnetic behavior.
(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 catalysts.
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 transitions are possible).
(iv) Because of their ability to 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 examples.
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, vanadium, 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 dichromate ?
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 reverberatory.
3FeO.Cr₂O₃+O₂ 2 Fe₂O₃+₄Cr₂O₃
[4Na₂CO₃+₂Cr₂O₃+3O_{2 4}Na₂CrO₄+₄CO₂]₂
4FeO.Cr₂O₃+ ₈Na₂CO₃+₇O₂₄Na₂CrO₄+₂Fe2O₃+O₂
The mass is then extracted with water, when sodium chromate is completely dissolved while Fe_2O3 is left behind.
(ii) Conversion of sodium chromate into sodium dichromate (NaCr₂O₇) : The sodium chromate extreme with water in previous steps is actifed.
3Na₂CrO₄+H₂SO₄Na₂Cr₂O₇+Na₂SO₄+H₂O
On cooling Na₂SO₄ separates out as Na₂SO₄. 10 H₂O and NaCr₂O₇ it remains in solution.
(iii) Conversion of NaCr2O7 into K₂Cr₂O₇ : The solution containing NaCr2O7 is treated with KCl.
NaCr₂O₇+KClK₂Cr₂O₇ +2NaCl 
Sodium chloride (NaCl) being less soluble separates out on cooling. On crystallizing the remaining solution, Orange coloured crystals of K₂Cr₂O₇ separate out.
Effect of Change of pH : When pH of solution of K₂Cr_2O7 is increased slowly as the medium changes from acidic to basic. The chromates and dichromates are interconvertible in aqueous solution depending upon pH solution.
Cr₂O_72- + H₂O   2CrO₄^2- + 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 orange coloured while at higher pH  (alkaline medium) it changes to yellow due to formation of chromate ions.

Q. 5. Describe the oxidizing 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 oxidizing and is used as primary standard in volumetric analysis involving oxidation of iodides, ferrous ion and S₂ -ions etc. In acidic solution its oxidizing action can be represented as follows :
Cr₂O₇^2- +₁₄H⁺+6e ₂Cr₃++7H₂O ; (E⁺=1.33V)

(a) It oxidizes potassium iodide to iodine.
Cr2O72- +14H +⁺⁶e - ₂Cr₃+₇H₂O+3I₂

(b) It oxidizes iron (II) salt to iron (III) salt
Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ + 2Cr³⁺ = 6Fe3 + +7H₂O

(c) It oxidizes H₂SO to S 
Cr₂O₇^2- + 8H⁺ + 3H₂S₂Cr₃+ +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 the presence of air : Pyrolusite ore is fused with alkali in the presence of air when potassium manganate 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 manganate and permanganate : The aqueous solution of is oxidized electrolytically or by using ozone or Cl₂ to obtain potassium permanganate. The process is carried out till Green color disappears and solution acquires distinet pink color.

MnO₄₂^- MnO₄^- + e^-  (oxidation at anode) (at anode) pink green color

H₂O+e-₁₂H₂+(OH)^- (reduction at cathode) (at cathode)

or, 2K₂MnO₄+Cl₂₂KMnO₄+₂KCl  
(green color)               (pink color)

Potassium permanganate is crystallized out from the solution.

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

(a) It oxidizes iron (II) salt to iron (III) salts.
2MnO₄^2- +16H⁺ +10Fe₂⁺ 2Mn₂⁺ +₈H₂O⁺ 10Fe₃⁺

(b) It oxidized sulfur dioxide to sulphuric acid.
2MnO₄^- + 5SO₂ + 2H2O5SO42- +2Mn2+ + 4H+

(c) It oxidizes oxalic acid to CO₂ and H₂O
2MnO₄^- + 16H⁺ +5C₂O₄₂Mn₂⁺ +8H₂^o + ₁₀CO₂

Q. 7. Use Hund’s rule to derive the electronic configuration of Ce₃⁺ and calculate its magnetic moment on the basis of the ‘spin only’ formula.
Ans⇒ The electronic configuration of Ce and Ce₃⁺ ions is :
Ce (Z=58)=54[Xe]4f1 5d1 6s₂
Ce₃⁺= 54[Xe]4f₁
The no. of unpaired electron =1
‘Spin only formula for magnetic moment of a species,
= n(n+2)B.M.
∴    Magnetic moment of Ce₃⁺
= n(n+2)B.M. = 3 B.M.=1.732 B.M.

Q. 8. Compare the chemistry of the actinides with that of lanthanoids with reference to (i) Electronic configuration (ii) Oxidation states and (iii) Chemical reactivity.
Ans⇒ (i) Electronic configuration : All the actinides 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 lanthanides are believed to have the electron configuration 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 have comparable energies and take part in giving different oxidation states.
Whereas in lanthanide  +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 actinides are highly reactive metals like lanthanides, 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 behavior.
(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 increased 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 compounds 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 sandwiched between s-block of the periodic table (highly reactive metals) and p-block (mostly non-metals). The last electron enters penultimate ((n-1)d) subshells in the atoms of these elements. These elements have partially filled d-subshells in their elemental states or in their commonly occurring oxidation states. Zn , Cd , Hg do not possess a partially filled inner. d-orbitals either in their elemental 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]₄f14 5d_{10 6}s₂         Hg (II)=5d10 
∴    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)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 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 , NiO , KMnO₄ , K₂Cr₂O₇ 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 lanthanide with special reference to :
(a) electronic configuration,
(b) atomic and ionic sizes,
(c) oxidation state and
(d) chemical reactivity.
Ans⇒ (a) Electronic 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 antepenultimate orbit in their elementary or ionic state. In both lanthanoids and actinoids the last electron enters (n-2)f subshell
Lanthanoids have general electronic configuration of [Xe]4f0-14 5d0-16s₂.
where [Xe] stands for Xe core.
Actinoids have the general electronic configuration of [Rn]5f0-14 6d0-27s₂ where (Rn) stands for radon core.
Last electrons enters 4f in the case of lanthanoids and it enters 5f in the case of actinides.
(b) Atomic sizes of Ln vary from 187 pm for La to 0.173 for Yb. Their ionic sizes [Ln3+] vary from 106 pm for La₃⁺ to 85 pm for Lu₃⁺, whereas the size of actinides 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] actinides show variable oxidations state of +2  , +3 , +4 , +5 , +6 , +7
(d) The chemical reactivity etc. are summarized below :
Comparison of Lanthanide and Actinide Series : Elements of lanthanide and actinides resemble in many respects but they do differ in some respects as shown below:

Q 12: To what extent do the electronic configurations decide the stability of oxidation states in the first series of transition elements? Illustrate your answer with examples.
Ans⇒ The oxidation states with the maximum number (+2 to +7) are exhibited by the elements of Mn oxidation states that are in the first half of the transition series. Ti  With the increase in atomic number, the stability of the +2 oxidation state increases. Once more electrons are filled in the d-orbital, this happens to a great extent.  The +2 oxidation state is not shown by the Sc. Its electronic configuration is 4s2 3d1. It loses all three electrons to form Sc3+ , +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar] ,Ti (+4) and V (+5) are very stable for the same reason. For Mn, the +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar]₃d₅.


Q.13: What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d3 , 3d5 ,3d8 and 3d4 ?
Ans⇒

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