NCERT Class 12 Chemistry Chapter 8 The d – and f – Block Elements

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NCERT Class 12 Chemistry Chapter 8 The d – and f – Block Elements

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The d – and f – Block Elements

Chapter: 8

1. Write down the electronic configuration of:

(i) Cr³⁺

Ans: Cr³⁺ (Chromium ion): 1s² 2s² 2p⁶, 3s² 3p⁶ 3d³

(ii) Cu⁺

Ans: Cu⁺ (Copper ion): 1s²,  2s², 2p⁶, 3s², 3p⁶, 3d¹⁰

(iii) Co²⁺

Ans: Co²⁺ (Cobalt ion): 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁷

(iv) Mn²⁺

Ans: Mn²⁺ (Manganese ion): 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵

(v) Pm³⁺

Ans: Pm³⁺ (Promethium ion): 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 5s², 5p⁶, 4f⁴

(iv) Ce⁴⁺

Ans: Ce⁴⁺ (Cerium ion): 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 5s², 5p⁶, 4f⁰

(vi) Lu²⁺

Ans: Lu²⁺ (Lutetium ion): 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 5s², 5p⁶, 4f¹⁴, 5d¹

(viii) Th⁴⁺

Ans: Th⁴⁺ (Thorium ion): 1s², 2s², 2p⁶, 3s², 3p⁶, d¹⁰, 4s², 4p⁶, 4d¹⁰, 5s², 5p⁶, 4f¹⁴, 5d¹⁰, 6s², 6p⁶

2. Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?

Ans: Mn²⁺ = 1s², 2s² 2p⁶, 3s², 3d⁵ (half filled d-orbitals)

Fe²⁺ = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁶

Mn²⁺ compounds are more stable due to their half-filled d-orbital configuration. Fe²⁺ compounds, on the other hand, are comparatively less stable as they have six electrons in their 3d-orbital. As a result, they tend to lose one electron (forming Fe³⁺) to achieve a more stable 3d⁵ configuration.

3. Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Ans: Mn²⁺ compounds are more stable due to their half-filled d-orbital configuration. Fe²⁺ compounds, on the other hand, are comparatively less stable as they have six electrons in their 3d-orbital. As a result, they tend to lose one electron (forming Fe³⁺) to achieve a more stable 3d⁵ configuration. In the second half of the first row transition series, electrons pair up in 3d orbitals. This increases the electronic repulsion.

4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Ans: When d orbitals are half-filled or completely filled, the oxidation state is more stable.

Eg. (i) ₂₅Mn = [Ar]3d⁵4s⁵

Mn²⁺ = [Ar] 3d⁵ (Most stable)

Mn³⁺ = [Ar] 3d⁵

Mm⁴⁺= [Ar] 3d³

Eg. (ii) ₂₉Cu= [Ar] 3d¹⁰4s¹

Cu⁺ = [Ar] 3d¹⁰ (Most stable)

Cu²⁺ = [Ar] 3d⁹

In manganese, the Mn²⁺ ion is more stable due to the symmetry and half-filled d-orbitals. Similarly, in copper, the Cu⁺ ion is more stable due to its symmetrical and fully filled d-orbitals.

5. 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: The stable oxidation state of transition element with the d electron configuration in ground state of atoms are as follows:

It should be noted that lower stable oxidation states generally result in the formation of ionic bonds, while higher oxidation states are typically associated with covalent bonds.

6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. 

Ans: [ScO₂]: Scandium (Sc) exhibits a +3 oxidation state (Group 3)

[MnO₄^–]^–: Manganese (Mn) in the +7 oxidation state (Group 7).

[CrO₄]²⁻​: Chromium (Cr) in the +6 oxidation state (Group 6).

Tio₃]^2-: Titanium (Ti) in the +4 oxidation state (Group 4).

[VO₃]^–: Vanadium (V) in the +5 oxidation state (Group 5).

[Cr₂O₇]^2-: Chromium (Cr) in the +6 oxidation state.

7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Ans: Lanthanoid contraction, also called lanthanide contraction, occurs as the atomic size or the ionic radii of the tripositive lanthanide ions steadily decrease from La to Lu because of the electrons entering the inner (n^-2) f orbitals and the increasing nuclear charge. This particular gradual decrease in the size with an increasing atomic number is referred to as lanthanide contraction. A group of fourteen elements following lanthanum i.e. from ₅₅ Ce 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 symbol ‘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 lanthanides) are of the form 4fn (n = 1 to 14 with increasing atomic number). These elements constitute one of the two series of inner transition elements or f- block.

Following are the consequences of lanthanide contraction:

(i) 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.

(ii) Cause of Lanthanide Contraction: The cause of lanthanide contraction is the increasing nuclear charge as one moves from ₅₈Ce to ₇₁Lu. With each added electron entering the inner 4f subshell, the poor shielding effect of these diffused orbitals allows the nucleus to exert a stronger pull on both 4f and outer electrons, reducing atomic size.

Consequences of Lanthanide Contraction:

(i) Similarly, elements of the second and third transition series, such as Sr and Hf, Nb and Ta, Mo and W, show a close resemblance in their properties. This similarity arises from their comparable sizes, which is influenced by the presence of lanthanides between them, causing lanthanide contraction.

(ii) Decrease in Basicity: With the decrease in ionic radii, covalent character of their hydroxides goes on increasing from Ce(OH), to Lu(OH), and so base strength goes on decreasing.

8. 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: Characteristics of the transition elements.

(i) Electronic configuration: General electronic configuration of these elements is (n-1)d^1-10 ng^1-2

(ii) Physical properties: These elements display metallic properties like lustre, high tensile strength, ductility, malleability, excellent thermal and electrical conductivity, hardness, and high melting points (except Zn, Cd, Hg).

(iii) Atomic and Ionic size: In a given series, there is a progressive decrease in radius with increasing atomic number.

(iv) Ionisation enthalpies: Ionization enthalpy increases across the series due to higher nuclear charge as d-orbitals are filled.

(v) Oxidation states: Transition elements exhibit multiple oxidation states.

Zinc (Zn), cadmium (Cd), and mercury (Hg) cannot be classified as true transition elements because their d-subshells are completely filled in both their elemental and common oxidation states. Transition elements typically have partially filled d-orbitals, which give rise to their unique chemical properties, unlike Zn, Cd, and Hg.

9. In what way is the electronic configuration of the transition elements different from that of the non transition elements?

Ans: Transition elements have incomplete penultimate d-orbitals, whereas the penultimate orbitals of representative elements (s- and p-block) are completely filled. The general valence shell configuration for s-block elements (Group 1 and 2) is ns¹⁻², and for p-block elements (Group 13 to 18) it is ns2np¹⁻⁶. In contrast, the valence shell configuration of transition elements is typically written as (n−1)d¹⁻⁹ns¹⁻².

10. What are the different oxidation states exhibited by the lanthanoids?

Ans: The most common oxidation state shown by lanthanoids is +3. In some exceptional cases it may be +2 or +4 (+2 in Eu and Yb; +4 in Ce).

11. Explain giving reasons:

(i) Transition metals and many of their compounds show paramagnetic behaviour.

Ans: Because transition metal atoms or ins have unpaired d-electrons in their configuration.

(ii) The enthalpies of atomisation of the transition metals are high.

Ans: Transition metals have high melting and boiling points because the atoms are held together by strong metallic bonds. Both the ns electrons and (n-1)d electrons participate in this bonding, making the metallic lattice very stable.

(iii) The transition metals generally form coloured compounds.

Ans: Transition metals are often coloured because their ions usually contain one or more unpaired electrons in their d-orbitals. This allows for d-d transitions, where an electron moves between d-orbitals, absorbing and emitting light, leading to colour.

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

Ans: 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 Haber’s process for NH₂ manufacture) and Ni (in catalytic hydrogenation) are examples of their good catalytic activities.

12. What are interstitial compounds? Why are such compounds well known for transition metals?

Ans: The compounds formed when small atoms of H, C or N get trapped inside the crystal lattice of metals is known as interstitial compounds.an trapped inside the crystal lattices of metais. They are generally non-stoichiometric and neither typically ionic or covalent.

Most of transition metals form interstitial compounds with small non-metal atoms such as hydrogen, boron, carbon and nitrogen. These small atoms enter into the void sites between the packed atoms of crystalline transition metals and form chemical bonds with transition metals. For example, steel and cast iron become hard by forming interstitial compounds with carbon.

The existence of vacant (n-1)d orbitals in transition elements and their ability to make bonds with trapped small atoms is the main cause of interstitial compound formation. Other examples are VH_0.56 TiH_1.7.

Some main characteristics of these compounds are:

(i) They exhibit high melting and boiling points, which surpass those of pure metals.

(ii) They are very hard. Some borides of transition elements approach diamond in hardness.

(ii) They are chemically inert but retain metallic conductivity.

13. How is the variability in oxidation states of transition metals different from

that of the non transition metals? Illustrate with examples.

Ans: In transition elements, the oxidation states change by a difference of one, resulting from the partial filling of d-orbitals.Mn exhibits +2, +3, +4, +5, +6 and +7 all differing by 1. In non-transition element, this variation is selective, always differing by 2 e.g., S exhibits 2, 4, 6 oxidation states, N exhibits 3, 5 etc.

14. Describe the preparation of potassium dichromate from iron chromite ore.

What is the effect of increasing pH on a solution of potassium dichromate?

Ans: The following steps are involved in preparation of K₂Cr₂O₇ from iron chromite (FeCr₂O₄ ) ore:

(i) Preparation of Sodium Chromate: The chromite ore ( FeO.Cr₂O₃ ) is finely powdered and mixed with sodium and quick lime and then heated to redness in a reverberatory furnace with free supply of air.

The solid is then treated with water, dissolving the sodium chromate completely, while Fe₂O₃ remains as a residue.

(ii) Conversion of Sodium Chromate into Sodium Dichromate (NaCr₂O₇): A solution of sodium chromate is filtered and then acidified to produce sodium dichromate:

3Na₂CrO₄ + H₂SO₄ → Na₂​Cr₂​O₇​ + Na₂​SO₄​ + H₂​O

The solution is subsequently cooled to precipitate crystallised sodium sulphate, while sodium dichromate remains dissolved.

(iii) Conversion of Na₂Cr₂O₇: Equimolar amounts of a hot solution of sodium dichromate and potassium chloride are combined to produce potassium dichromate:

Na₂​Cr₂​O₇​ + 2KCl → K₂​Cr₂​O₇ ​+ 2NaCl

Sodium chloride (NaCl) being less soluble separates out on cooling. On crystallising the remaining solution, orange coloured crystals of K₂Cr₂O₇ separate out.

(iv) 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 upon pH of solution.

In a low pH (acidic medium), a solution of K₂Cr₂O₇ appears orange, while at higher pH (alkaline medium), it turns yellow due to the formation of chromate ions..

15. Describe the oxidising action of potassium dichromate and write the ionic

equations for its reaction with:

(i) iodide. 

(ii) iron(II) solution. and 

(iii) H₂S.

Ans: Potassium dichromate (K₂Cr₂O₇) is a powerful oxidising agent used in volumetric analysis to oxidise iodides, ferrous ions, and sulphide ions. In acidic solution, its reduction is represented by: 

Cr₂O₇²⁻ + 14H⁺+6e− → 2Cr³⁺ + 7H₂O;

with an E° value of 1.33V.

(a) It oxidises potassium iodide to iodine.

Cr₂O₇^2- + 14H + 6I^– 2Cr ₃ + 7H₂O + 3I₂

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

Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ 2Cr³⁺ + 6Fe³⁺ + 7H₂O

(c) It oxidises H2S to S

Cr₂O₇^2- +8H⁺ +3H ₂ S 2Cr³⁺ + 7H₂O + 3S

16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with

(i) iron(II) ions.

(ii) SO₂. and 

(iii) oxalic acid?

Write the ionic equations for the reactions.

Ans: Preparation of KMnO, from pyrolusite ore (MnO₂) involves the following steps:

Oxidation of Manganese Dioxide (MnO₂): Manganese dioxide is fused with potassium hydroxide (KOH) and heated in the presence of an oxidizing agent, such as air or potassium nitrate (KNO₃). This produces potassium manganate (K₂MnO₄):

2MnO₂​ + 4KOH + O₂​ → 2K₂​MnO₄ ​+ 2H₂​O

(ii) Oxidation of Manganate to Permanganate: The aqueous solution of K₂MnO₄ is oxidized either electrolytically or by using ozone or chlorine (Cl₂) to obtain potassium permanganate. The process continues until the green colour vanishes, and the solution turns a distinct pink colour.

Potassium permanganate is crystallised out from the solution.

Oxidising Properties: It acts as a powerful oxidising agent in different media differently. In acidic medium, it oxidises iron(II) salts to iron(III) salts, 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⁺ + 8H₂​O + 10Fe³⁺

(b) It oxidised sulphur dioxide to sulphuric acid.

2MnO₄⁻​ + 5SO₂ ​+ 2H₂​O → 5SO₄²⁻ ​+ 2Mn²⁺ + 4H⁺

(c) It oxidises oxalic acid to CO₂ and H₂O

2MnO₄⁻​ + 16H⁺ + 5C₂​O₄²⁻​ → 2Mn²⁺ + 8H_2​O + 10CO₂​

17. For M²⁺/M and M³⁺/M²⁺ systems the E° values for some metals are as follows:

Use this data to comment upon:

(i) the stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺ and

(ii) the ease with which iron can be oxidised as compared to a similar process

for either chromium or manganese metal.

Ans: (i) E°_(cr3+/cr2+) is negative (-0.4V). It shows that stability of Cr³⁺ ions, i.e. Cr³⁺ in solution cannot be reduced to Crions. Further Mn³⁺ has high positive E° value so it is easily converted to Mn²⁺ as compared to the conversion of Fe³⁺ into Fe²⁺. Thus, the order of relative stabilities of different ions is:

Mn³⁺ Fe³⁺ < Cr³⁺

(ii) Mn²⁺/Mn has most negative reduction potential and it is most easily oxidised. The ease of oxidation is Mn > Cr > Fe.

18. Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.

Ans: Transition metal ions with incompletely filled d orbitals are coloured and  Ions which has complete or vacant d-orbitals are colourless.

Since Sc³⁺ and Cu⁺ have 3d⁰ and 3d¹⁰ configurations, their aqueous solutions are colourless. In contrast, Ti³⁺, V³⁺, Mn²⁺, Fe³⁺, and Co²⁺ have coloured aqueous solutions.

19. Compare the stability of +2 oxidation state for the elements of the first transition series.

Ans: In the beginning of 3d transition series, Se²⁺ is virtually not known or in other words it is not stable in comparison to Sc³⁺, Ti³⁺, V³⁺, Cr²⁺ are known but less stable in comparison to their most common oxidation state of +3.

In the middle of the series, Mn²⁺, Fe²⁺, and Co²⁺ are more common. Mn²⁺ and Mn⁷⁺ are particularly stable oxidation states for manganese. Fe²⁺ is less stable compared to Fe³⁺ because Fe³⁺ achieves a half-filled d⁵ configuration, which provides extra stability. Co²⁺ is also less stable compared to Co³⁺. However, Ni²⁺ is the most stable among its possible +2, +3, and +4 oxidation states.Towards the end of the series, Cu²⁺ is more stable and commonly found compared to Cu⁺. Lastly, Zn forms only the highly stable Zn²⁺ due to its fully filled 3d¹⁰ configuration.  

20. Compare the chemistry of actinoids with that of the lanthanide with special reference to:

(i) electronic configuration. 

(ii) atomic and ionic sizes. and 

(iii) oxidation state.

(iv) chemical reactivity.

Ans:

21. How would you account for the following: 

(i) Of the d 4 species, Cr²⁺ is strongly reducing while manganese(III) is strongly oxidising. 

Ans: Cr²⁺, with a d⁴ configuration, is a strong reducing agent that oxidises to Cr³⁺, which has a d³ configuration.This d3 configuration can be written as configuration, which is a more stable configuration. In the case of Mn³⁺ (d⁴), it acts as an oxidising agent and gets reduced to Mn²⁺ (d⁵). This has an exactly half-filled d-orbital and is highly stable.

(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. 

Ans: This d³ configuration can be written as  configuration, which is a more stable configuration. In the case of Mn³⁺ (d⁴), it acts as an oxidising agent and gets reduced to Mn²⁺ (d⁵). This has an exactly half-filled d-orbital and is highly stable. Co(II) = [Ar] 483d7 and Co(III) = [Ar]₁, 45°3d 16 18 In Co (III) specie, 6 lone pairs of electrons from ligands are accommodated by sp’d hybridisation which is not possible in Co (II).

(iii) The d¹ configuration is very unstable in ions. 

Ans: Some species with d’ configuration are reducing and tends to loose one electron to acquire d’ stable configuration. Some other species with d’ configuration like Cr (V) and Mn(VI) undergo disproportionation.

22. What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution. 

Ans: Disproportionation represents a special type of redox reaction in which a reactant is both oxidized and reduced during the course of a reaction. Disproportionation reactions can be identified by looking for an element with three oxidation states within the same reaction. 

In other words , same substance is behaving as an oxidising agent for one molecule and at the same time as a reducing agent for another molecule.

E.g.

(i) 

(ii) 

23. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? 

Ans: Cu = [Ar]3d¹⁰4s’: When copper atom loses 1 electron, it exhibits +1 oxidation state and forms C u⁺ ion with stable 3d¹⁰ configuration. Copper metal in the first series of transition metals exhibits +1 oxidation state most frequently as it readily loses one electron (present in 4s orbital) to give stable 3d¹⁰ electronic configuration.

24. Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution? 

Ans: Mn³⁺ (4 unpaired), Cr³⁺ (3 unpaired), V³⁺ (2 unpaired), Ti³⁺ (1 unpaired). Cr³⁺ is the most stable in aqueous solution due to its half-filled t₂g (d³) configuration.

Out of these species, Cr³⁺ is most stable in aqueous solution due to its tendency of complex formation.

25. Give examples and suggest reasons for the following features of the transition metal chemistry: 

(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic. 

Ans: In the case of a lower oxide of a transition metal, the metal atom has some electrons present in the valence shell of the metal atom that are not involved in bonding. As a result, it can donate electrons and behave as a base. In the case of a higher oxide of a transition metal, the metal atom does not have an electron in the valence shell for donation. As a result, it can accept electrons and behave as an acid.

Example: MnO is a basic oxide, while Mn₂O₇ is an acidic oxide.

(ii) A transition metal exhibits the highest oxidation state in oxides and fluorides.

Ans: It is due to the fact that fluorine and oxygen are the two elements which are most electronegative in the Periodic Table. For example, in OsF6 and V₂O₅, the oxidation states of Os and V are +6 and +5 respectively.

(iii) The highest oxidation state is exhibited in oxoanions of a metal. 

Ans: Oxygen is a strong oxidising agent due to its high electronegativity and small size.So, oxo-anions of metal have the highest oxidation state.For example, in MnO⁻⁴, the oxidation state of Mn is +7. The highest oxidation state of transition element is seen in oxides and these oxides are acidic in nature. These acidic oxides dissolve in base and form oxoanions of transition metal. Thus oxoanions of transition element show highest oxidation state of transition metal.

26. Indicate the steps in the preparation of: 

(i) K₂ Cr₂ O₇ from chromite ore. 

Ans: The preparation of potassium dichromate (K₂Cr₂O₇) from chromite ore involves the following main steps:

(a) The chromate ore is finely ground and heated strongly with molten alkali in the presence of air. 

2FeCr₂O, + 8NaOH + 7/20, 4Na, Cr₂O, + Fe₂O, + 4H,0 

Chromite Sodium chromate

(b) The sodium chromate solution is filtered and acidified with dilute sulfuric acid, yielding sodium dichromate as a result.

2Fe Cr₂O + H₂SO₄ → Na, Cr₂O, + Na₂SO₄ + H₂O 

Sodium chromate Sodium dichromate

(c) A calculated quantity of potassium chloride is added to a hot concentrated solution of sodium dichromate. Potassium dichromate is less soluble therefore it crystallises out first.

Na,Cr₂O, + 2KCl → K₂Cr₂O₂ + 2NaCl 

Sodium dichromate Potassium dichromate

(ii) KMnO₄ from pyrolusite ore. 

Ans: (a) Pyrolusite ore is fused with an alkali and air, resulting in the formation of potassium manganate during the reaction.

2MnO₂ + 4KOH + O₂ → 2KMnO₄ + 2H₂O 

Potassium manganate

(b) Potassium manganate is oxidised by using either CO₂, ozone or chlorine to potassium 2 permanganate.

2KMnO + Cl₂ → 2KMnO₄ + 2KCl₄

Potassium permanganate

(c) Potassium permanganate is crystallised by cooling its concentrated solution, allowing KMnO₄ crystals to form, which are then filtered.

27. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. 

Ans: 1. A metal alloy is a substance that combines more than one metal or mixes a metal with other non-metallic elements.

2. An important alloy containing lanthanoid metal (~95%), iron (~5%), traces of S, C, Ca and Al is mischmetal.

3. Mischmetal is used in Mg-based alloy to produce bullets, shells and lighter flint.

28. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104. 

Ans: Inner transition elements are a group of elements in the periodic table that are also known as f-block elements. They generally belong to group 3 in the periodic table but are mentioned separately as the f block elements. These f block elements are known as inner transition elements.

The electronic configuration of given atomic number are as:

29 = [Ar] 4s¹3d¹⁰ (d-block element or transition element)

59 = [Xe] 6s²4f³ (f-block element or transition element)

74 = [Xe] 6s¹4f¹⁴5d⁵ (d-block element or inner transition element)

95 = [Rn]7s²5f⁷6d° (f-block. element or inner transition element)

102 = [Rn] 7s²5f¹⁴ (f-block element or inner transition element)

104 = [Rn] 7s²5f⁴6d² (d-block element) 

Thus, atomic numbers 59 (lanthanoid), 95 and 102 (actinoids) are included among inner transition elements.

29. The chemistry of the actinide elements is not so smooth as that of the lanthanides. Justify this statement by giving some examples from the oxidation state of these elements. 

Ans: Among the actinides, there is a greater range of oxidation states as compared to lanthanoids. Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium exhibit oxidation states of +3, +4, +5, and +6, while neptunium shows +3, +4, +5, and +7. The most common oxidation state among actinoids is +3.

30. Which is the last element in the series of the actinides? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Ans: The general electronic configuration of actinides is [Rn] 5f1-14 6d0-1 7s2. Here, [Rn] is the electronic configuration of the nearest noble gas, which is radium.

Lawrencium (Lr: Z = 103) is the last element in the actinide series.

Configuration₁₀₃ Lr = [Rn] 5f ¹⁴6d ¹7s² 

Its possible oxidation state = + 3.

31. Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula. 

Ans: Electronic Configuration of Cerium (Ce)

Ce = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹5d¹6s²

Ce(Z=58) = ₅₄[Xe]4f¹ 5d¹6s²

Electronic Configuration of Ce³⁺: 

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹

Ce³⁺ = ₅₄[Xe]4f¹

The number of unpaired electron = 1

Using the ‘spin-only’ formula:

μ = √n(n+2)​ BM

∴ Magnetic Moment of Ce³⁺

μ = √1(1+2)

√3​ = 1.73 BM

32. Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements. 

Ans: + 4 oxidation states exist in₅₈ Ce, ₅₉ Pr and ₆₅Tb.

+2 oxidation states exists in Nd, ₆₂ Sm ₆₅ Eu ₆₉Tm and 70^{(1 / b)}

Generally +2 oxidation state is exhibited by the elements with configuration 3d deg 6s² sol. that two electrons may be easily lost. Similarly, the +4 oxidation state is exhibited by elements that, after losing four electrons, achieve a configuration close to 4fⁿ or 4f⁷.

33. Compare the chemistry of the actinides with that of lanthanoids with reference to: 

(i) Electronic configuration.

Ans: All the actinides are believed to have the electron configuration of 7s² and variable occupancy of the 5f and 6d 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 states. and 

Ans: 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.

In lanthanide, the +3 oxidation state is predominant because the 6s and only one 4f orbital participate in bonding. However, +2 and +4 oxidation states can occasionally be observed in solutions or solid compounds.

(iii) Chemical reactivity. 

Ans: 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.

34. Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109. 

Ans: Following are the electronic configurations of the elements: 

Electronic Configuration(61): [X_a] 4f⁵6s²

Electronic Configuration(91): [R_n] 5f²6d¹7s²

Electronic Configuration(101): [R_n] 5f¹³7s²

Electronic Configuration(109): [Rn] 5f¹⁴6d⁷7s²

35. Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: 

(i) electronic configurations 

Ans: In the first transition series, the 3d orbitals are progressively filled, while in the second and third transition series, the 4d and 5d orbitals are filled, respectively, with similar electronic configurations.

(ii) oxidation states 

Ans: The elements in the same vertical column generally show similar oxidation states. The number of oxidation states shown by the elements in the middle of each series is maximum and minimum at the extreme ends.

(iii) ionisation enthalpies. and 

Ans: The first ionisation enthalpies in each series generally increase gradually from left to right, although some exceptions occur. In the second (4d) series, certain elements have higher, while others have lower, ionisation enthalpies compared to their counterparts in the 3d series within the same vertical column.

(iv) atomic sizes. 

Ans: The 4d series elements have larger atomic sizes compared to the 3d series. However, the 4d and 5d series elements exhibit nearly the same atomic size due to the effect of lanthanide contraction.

36. Write down the number of 3d electrons in each of the following ions: Ti ²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺. Indicate how you would expect the five 3d orbitals to be occupied for these hydrated ions (octahedral). 

Ans: The quantity of 3d electrons and the arrangement of 3d orbitals for various transition metal ions are outlined as follows.

37. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements. 

Ans: The heavier transition elements belong to the fourth (4d) and fifth (5d) and sixth (6d) transition series. Their properties are expected to be different from the elements belonging to the first (3d) series due to the following reasons:

(i) The atomic radii of 4d and 5d series elements are larger due to more electron shells, but the difference between them is smaller due to lanthanoid contraction.

(ii) Because of stronger interatomic binding, the m.p and b.p of the elements of 4d and 5d series are higher.

(iii) The enthalpies of atomisation for elements in the first transition series are lower compared to those in the second and third transition series due to weaker metallic bonding in the former series.

(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of strong metallic bonding (M-M bonding)

Actually, the atomic size decreases due to this, and the effective nuclear charge increases. As a result, there is an increase in ionisation energy among the 3d elements.

38. What can be inferred from the magnetic moment values of the following complex species?

Ans: Magnetic moment (μ) is given as  μ = √n (n+2)

For n = 1,μ = √1(1 + 2) = √3 = 1.73BM

For n = 2, μ = √2 × (2 + 2) = √8 = 2.83 BM

For n = 3, μ = √3 × (3 + 2) = √15 =3.87 BM

For n = 4, μ = √4 × (4 + 2) = √24 = 4.899 BM

For n = 5, μ = √5 × (5 + 2) = √35 = 5.92 BM

K₄[Mn(CN)₆], 

Mn²⁺ : 3d⁵

The magnetic moment of 2.2 BM indicates 1 unpaired electron, suggesting a low-spin or inner orbital complex with the configuration t⁵_2g, where electron pairing occurs due to strong field ligands.

[Fe(H₂O)₆]²⁺, 

Fe²⁺ : 3d⁶

The magnetic moment of 5.3 BM indicates 4 unpaired electrons, suggesting a high-spin or outer orbital complex with the configuration t⁴_2ge²_g​, typical for weak field ligands.

K₂[MnCl₄], 

Mn²⁺ : 3d⁵

In this compound, Mn is in +2 state (Mn²⁺), μ = 5.9 BM shows the presence of five unpaired electrons. Hence, the hybridization involved will be sp³ and the complex will be tetrahedral.