The d-and f-Block Elements PDF

# The d-and f-Block Elements ## Explain why size of lanthanoids decreases from La to Lu? **Lanthanoid Contraction:** In the lanthanoid series, with increasing atomic number, the atomic and ionic radii decrease from one element to another but the decrease is very small. For example, on moving from...

# The d-and f-Block Elements ## Explain why size of lanthanoids decreases from La to Lu? **Lanthanoid Contraction:** In the lanthanoid series, with increasing atomic number, the atomic and ionic radii decrease from one element to another but the decrease is very small. For example, on moving from Ce to Lu, the atomic radii decreases from 183 pm to 173 pm and the decrease in only 10 pm. Similarly, the ionic radii decreases 103 pm to 85 pm on moving from Ce³⁺ to Lu³⁺ ions and the decrease is only 18 pm. Thus, for an increase of atomic number 14, the decrease in atomic radii or ionic radii are very small, only 10 pm and 18 pm respectively. This is a very small decrease in comparison to elements of other groups and periods. The steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic number is called lanthanoid contraction. **Cause of Lanthanoid Contraction:** In the lanthanoid series, as we move from one element to another, the nuclear charge increases by one unit and one electron is added. The new electrons are added to the same inner 4f-subshells. However, the 4f-electrons shield each other from the nuclear charge poorly because of the very diffused shapes of the f-orbitals. The nuclear charge, however, increases by one at each step. Hence, with increasing atomic number and nuclear charge, the effective nuclear charge experienced by each 4f-electron increases. As a result, the whole of 4f-electron shell contracts at successive element, though the decrease is very small. This result gradual decrease in size of lanthanoids with increases in atomic number. The sum of the successive reductions gives the total lanthanoid contraction. ## Consequences of Lanthanoid Contraction The important consequences of lanthanoid contraction are: 1. **Resemblance of Second and Third Transition Series:** It has significant effect on the relative properties of the elements before and after the lanthanoids in the periodic table. There is a regular increase in size from Sc to Y and Y to La. Similarly, we expect normal increase in size in group as: Ti - V - Zr - Nb - Hf - Ta, etc. However, after the lanthanoids, the increases in size from second to third transition series almost vanishes. The pairs of elements: Zr-Hf, Nb-Ta, Mo-W, etc. possess almost the same size. The properties of these elements are also very similar. As a result of lanthanoid contraction, the elements of second and third transition series resemble each other much more than the elements of first and second transition series. 2. **Similarity Among Lanthanoids:** Because of the very small change in radii of lanthanoids their chemical properties are quite similar. Thus, it is difficult to separate the elements in pure form. Recently, methods based on repeated fractional crystallization or ion exchange techniques, with the advantage of slight differences in their properties (like solubility, complex ion formation, hydration, etc.) arising from very slight size difference in trivalent ions have been used. 3. **Basicity Differences:** Due to lanthanoid contraction, the size of lanthanoid ions decreases regularly with increase in atomic number. Because of decrease in size, their covalent character between lanthanoid ion and OH ions increases from La³⁺ to Lu³⁺. Therefore, the basic strength of the hydroxides decreases with increase in atomic number. La(OH)3 is most basic while Lu(OH)3 is the least basic. ## Long Answer Type Questions 1. Explain giving reasons: - Transition metals or ions and many of their compounds show paramagnetic behavior. - The enthalpies of atomisation of the transition metals is high. - The transition metals generally form coloured compounds. - Transition metals and their many compounds act as good catalysts. **Ans.** - Most of the transition elements and their compounds show paramagnetic behaviour due to the presence of unpaired electrons in atoms, ions or molecules. For example: Cu²⁺ : 1s² 2s² 2p⁶ 3s²3p⁶ 3d⁹ or 1s², 2s²2p⁶, 3s²3p⁶ 11 11 11 11 1 Cu²⁺ has one unpaired electron, therefore it is paramagnetic in nature. - They have strong interatomic interaction and strong bonding between the atoms. It is due to the presence of a large number of unpaired electrons in their atoms. Therefore, transition elements have high enthalpies of atomisation. - Refer to Short Answer Type Question (II) No. 2. - Transition metals and their compounds, particularly oxides, have good catalytic properties. Catalytic properties of these metals are due to the presence of vacant d-orbitals and the tendency to exhibit variable oxidation state. 2. How many types of oxidation state show by the lanthanoids? **Ans.** Lanthanoids exhibit limited number of oxidation states because the energy difference between 4f and 5d-subshells is large. - **Ce and Tb Exhibit +4 Oxidation State:** Cerium (Ce) and Terbium (Tb) attain f⁰ and f⁷ configurations respectively when they go in +4 oxidation state as shown below: Ce⁴⁺: [Xe]4f⁰ Tb⁴⁺: [Xe]4f⁷ - **Eu and Yb Exhibit +2 Oxidation State:** Europium and Ytterbium get f⁷ and f¹⁴ configurations in +2 oxidation state as shown below: Eu²⁺: [Xe]4f⁷ Yb²⁺: [Xe]4f¹⁴ - **La, Gd and Lu exhibit only +3 Oxidation States:** These elements show +3 oxidation states only because by losing three electrons they acquire a stable configuration of either half-filled and completely filled 4f-subshells. The stability of different oxidation states has significant effect on the properties of these elements. For example, Ce (IV) is favoured because of its gas configuration. But it is a strong oxidising agent changing to common +3 oxidation state. As already seen, Pr, Nd, Tb and Dy also show +4 oxidation states but only in oxides MO₂. Similarly Eu²⁺ is stable because of its half-filled 4f⁷ configuration. However, it is a strong reducing agent changing to Eu³⁺ (common oxidation state). Similarly, Yb²⁺ having configuration 4f¹⁴ is a reductant. Tb also has half-filled f-orbitals and is an oxidising agent. - **Exceptional Examples:** Some other elements show +2 and +4 oxidation states even though they have electronic configurations other than f⁰, f⁷, f¹⁴. For example, Sm²⁺(4f¹³), Pr⁴⁺(4f¹), Tm²⁺(4f¹²), Dy⁴⁺(4f⁸), Nd⁴⁺(4f²), etc. But these states are less stable than the +3 state which characteristic of this family. 3. What is lanthanoid contraction? What are the consequences of lanthanoid contraction? Describe lanthanoid contraction and its effects on the metallic character. **Or** What is lanthanoid contraction? Explain its reason **Ans.** Lanthanoid contraction is the steady decrease in size of lanthanoids with increase in atomic number. The reason for this is that, the new electrons are added to the inner 4f-subshells, and the 4f-electrons shield each other from the nuclear charge poorly because of the very diffused shapes of the f-orbitals. As the nuclear charge increases by one unit at each step, the effective nuclear charge experienced by each 4f-electron increases, leading to the contraction of the 4f-subshell. The sum of the successive reductions gives the total lanthanoid contraction. The important consequences of lanthanoid contraction are: - There is a regular increase in size from Sc to Y and Y to La in the periodic table. However, from second to third transition series, the size increase is almost negligibly small, due to the lanthanoid contraction. - Lanthanoids have very similar chemical properties due to the small change in their size. This makes separation of lanthanoids a difficult task. - The basic strength of lanthanoid hydroxides decreases with increase in atomic number due to the increase in covalent character with decreasing size of the lanthanoid ions. ## Short Answer Type-II Questions 1. Which elements are called transition elements? Explain. Write their two main characteristics and their general configuration. **Or** What are transition elements? Write the name and electronic configuration of the element of the first transition series. **Ans.** The elements having incomplete (n-1)d-orbital in the ground state or in any of its excited states are called transition elements. These are also called d-block elements. For example: - Cr (24): 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵, 4s¹ - Cu (29): 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s¹ The transition elements are classified into four series i.e., - 3d series - 4d series - 5d series - 6d series Characteristics of Transition Elements: - Transition elements are metals and good conductors of heat and electricity. - Transition elements show the variable valencies due to the participation of (n - 1)d and ns electrons. - Transition metals are less reactive than s-block elements due to high ionisation potential. - Transition metals and their compounds act as catalysts due to the presence of unpaired d-electrons. - Their general electronic configuration is: (n-1)d¹-¹⁰ ns¹-² The electronic configuration of the first transition series is Sc: [Ar]3d¹4s² 2. Explain the formation of interstitial compounds. Describe four special characteristics of transition elements. **Or** Explain the formation of interstitial compounds? **Or** Write electronic configuration of Cr (Z = 24) and Cu (Z=29) and also explain the two main characteristics of transition elements. **Ans.** Formation of Interstitial Compounds: Transition elements are able to entrap smaller atoms of nonmetal elements like C, H and N in interstitial sites of a crystal lattice. Such compounds are called interstitial compounds. In these compounds, the trapped atoms get bonded to the atoms of transition element in such a way that interstitial compounds are formed. Four Special characteristics of transition elements: - Transition elements are metals and good conductors of heat and electricity. - Transition elements show the variable valencies due to the participation of (n - 1)d and ns electrons. - Transition metals are less reactive than s-block elements due to high ionisation potential. - Transition metals and their compounds act as catalysts due to the presence of unpaired d-electrons. The electronic configuration of: - Cr (24): [Ar]3d⁵4s¹ - Cu (29): [Ar]3d¹⁰4s¹ The two main characteristics of transition elements are: - Variable valencies due to the participation of (n - 1)d and ns electrons. - Formation of interstitial compounds and coloured ions by transition elements. 3. What are transition elements? **Or** Explain transition metal generally form coloured compounds. **Ans.** The elements having incomplete (n-1)d-orbital in the ground state or in any of its excited states are called transition elements. These are also called d-block elements. The colour of transition metal compounds is due to the presence of unpaired electrons in (n-1)d-orbital. For example: Cu²⁺(Blue), Co²⁺(Blue), Ni²⁺(Green). 4. Write the electronic configuration of lanthanides and actinoids elements. **Ans.** There are two series of f-block elements 4f- and 5f-series. Each series there are 14 elements. Their general electronic configuration is as follows : - **Lanthanoids** = [xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s² - **Actinoids** = [Rn] 5f¹⁻¹⁴ 6d⁰⁻¹ 7s² (except thorium) 5. Write the formulae of the following : - White vitriol - Blue vitriol - Green vitriol - Mohr's salt **Ans.** - **White vitriol:** ZnSO₄.7H₂O - **Blue vitriol:** CuSO₄.5H₂O - **Green vitriol:** FeSO₄.7H₂O - **Mohr's salt:** (NH₄)₂Fe(SO₄)₂.6H₂O 6. Explain why? - Scandium (Z = 21) is a transition element while Zn (Z = 30) is not. - Cr²⁺ is reducing agent while Mn³⁺ is an oxidising agent though both have same d-configuration. **Ans.** - Sc (21) = [Ar] 4s²3d¹ - Zn (30) = [Ar] 4s²3d¹⁰ Scandium has one unpaired electron in its 3d-orbital while Zn does not have any, hence Zn is not considered as a transition element. Cr²⁺ is a reducing agent as its configuration changes from d⁴ to d³, the latter having a half-filled t₂g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in half-filled d configuration which has extra stability. 7. Write the preparation of potassium dichromate (K₂Cr₂O₇ ) and potassium permanganate (KMnO₄). Discuss the method of preparation of potassium permanganate from pyrolusite and write the structural formula of permanganate ion. Discuss the method of preparation of potassium dichromate from chromite ore and write the structural formula of dichromate ion. [2024] **Ans.** * **Preparation of potassium dichromate (K₂Cr₂O₇ )**: K₂Cr₂O₇ is prepared from chromite ore, (FeCr₂O₄ or FeO.Cr₂O₃). Iron (II) chromite. It is prepared in the following three steps: **Step 1: Conversion of chromite ore into sodium chromate:** Concentrated chromite ore is fused with sodium hydroxide or sodium carbonate in the presence of air. 4FeCr₂O₄ + 16NaOH + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8H₂O or 4FeO Cr₂O₃ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ **Step 2: Conversion of sodium chromate into sodium dichromate:** The yellow solution of sodium chromate is filtered and acidified with conc. H₂SO₄, giving sodium dichromate. 2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O On cooling sodium sulphate crystallizes out first as Na₂SO₄ 10H₂O and removed. Remaining solution contains sodium dichromate. **Step 3: Conversion of sodium dichromate into potassium dichromate**: Potassium dichromate is prepared by mixing a hot concentrated solution of sodium dichromate and potassium chloride in equimolar proportions. Na₂Cr₂O₇ + 2KCI → K₂Cr₂O₇ + 2NaCl Sodium chloride is precipitated out from hot solution and removed by filtration. Orange crystals of potassium dichromate separate out from mother liquor on cooling. * **Preparation of Potassium Permanganate (KMnO₄):** - **Step 1: Conversion of pyrolusite to potassium manganate:** Pyrolusite is fused with potash (caustic) or potassium carbonate in the presence of air or potassium nitrate. Green coloured potassium manganate is formed. 2MnO₂ + 4KOOH + O₂ → 2K₂MnO₄ + H₂O or 2MnO₂ + K₂CrO₃ + O₂ →2K₂MnO₄ + 2CO₂ or MnO₂ + 2KOH + KNO₃ → K₂MnO₄ + KNO₂ + H₂O - **Step 2: Oxidation of potassium manganate to potassium permanganate:** The green mass obtained in the first step is extracted with water which gives a green solution of potassium manganate. The solution is treated with chlorine or ozone or carbon dioxide current to oxidise potassium manganate to potassium permanganate. The voilet coloured solution is concentrated and dark purple crystal of KMnO₄ separate out. 2K₂MnO₄ + Cl₂ → 2KMnO₄ + 2KCl or 2K₂MnO₄ + O₃ + H₂O → 2KMnO₄+ 2KOH + O₂ or 3K₂MnO₄ + 2CO₂→2MnO₂ + 2K₂CO₃ + 2KMnO₄ In the oxidation of MnO₄²- manganate ion disproportionates in an acidic or neutral medium and give permangnate MnO₄⁻ ion. 3MnO₄²⁻ + 4H⁺→ 2MnO₄⁻ + MnO₂ + 2H₂O - **Step 3: Commercial preparation :** Commercially KMnO₄ is prepared by electrolytic method. The alkaline potassium manganate solution is electrically oxidised. The potassium manganate solution is taken in an electrolytic cell having nickel anode and iron cathode. Potassium manganate solution is in the anodic compartment while dil alkali (KOH) solution is added in the cathodic compartment. On passing the current manganate ion is oxidised to permanganate ion at anode and hydrogen is liberated at cathode. 2K₂MnO₄ → 2K⁺ + MnO₄²⁻ At cathode: 2H⁺ + 2e⁻ → 2H From water At anode : 2H → H₂ Purple MnO₄²⁻ → MnO₄⁻ + e⁻ Green **Structure of Permanganate ion** **Structural formula of dichromate ion** ## Short Answer Type-I Questions 1. Why does transition elements show variable oxidation states? **Ans.** General electronic configuration for transition elements is (n-1)d¹⁻¹⁰ ns¹⁻². There is very little energy difference between (n-1)d and ns orbitals. Thus in transition elements both orbitals (n - 1)d and ns readily forms the bond. ns electrons contribute in + 1 and + 2 oxidation state while for higher oxidation states; e. g., +3, +4, +5,+ 6, etc. both ns and (n-1)d orbitals contributes. All the element of the first transition series except Sc show an oxidation state of + 2. This can be attributed to loss of two outer 4s electrons and oxidation state of + 3 to + 7 is due to the participation of two 4s and one-five 3d-electrons in bond formation. 2. Explain, why most of the transition metals form coordination compounds and exhibit variable valency. **Ans.** Due to small size of transition metal cation, high effective nuclear charge and availability of vacant d-orbitals for bond formation transition elements show a tendency to form complex compounds. Variable oxidation number means that elements show many oxidation states in its compounds. In each series, except first and last element, rest all the transition elements show variable oxidation state. This is because there is very less energy difference between ns-orbital and (n - 1) d-orbital thus both the orbitals participate in bond formation. As an exception Sc and Zn show only one oxidation state, i.e., +3 and +2 respectively. 3. Write general electronic configuration of transition metal and explain shielding effect. **Ans.** Transition series: The d-block elements are composed of four complete rows of ten elements each. These rows are called first, second, third and fourth transition series. These series involve the filling of 3d, 4d, 5d and 6d orbitals respectively. The genera electronic configuration of d-block elements is : - [noble gas configuration] (n - 1) d¹⁻¹⁰ns¹⁻² - e.g., the electronic configuration of 21Sc is: 1s², 2s²2p⁶, 3s²3p⁶3d², 4s² or [18Ar] 3d 4s² Interelectronic forces and variation in total nuclear charge plays a vital role to decide the electronic configuration of ions because 4s-orbital is filled first. 4. Write the formula of the following: - White vitriol - Blue vitriol - Green vitriol - Mohr's salt **Ans.** - White vitriol: ZnSO₄.7H₂O - Blue vitriol: CuSO₄.5H₂O - Green vitriol: FeSO₄.7H₂O - Mohr's salt: (NH₄)₂Fe(SO₄)₂.6H₂O 5. Explain why ? - Scandium (Z = 21) is a transition element while Zn (Z = 30) is not. - Cr²⁺ is reducing agent while Mn³⁺ is an oxidising agent though both have same d-configuration. **Ans.** - Sc (21) = [Ar] 4s²3d¹ - Zn (30) = [Ar] 4s²3d¹⁰ Scandium has one unpaired electron in its 3d-orbital while Zn does not have any, hence Zn is not considered as a transition element. Cr²⁺ is a reducing agent as its configuration changes from d⁴ to d³, the latter having a half-filled t₂g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in half-filled d configuration which has extra stability.