The d and f Block Elements

The chapter “The d- and f-Block Elements” from the 12th standard chemistry part 1 NCERT Board textbook explores the elements occupying the middle and bottom sections of the periodic table. Here’s a summary of the key concepts:

1. General Introduction:

• d-Block Elements (Transition Elements): Located in groups 3-12 of the periodic table, these elements have their outermost electron(s) in the d subshell. They exhibit a transition in properties between the highly reactive s– and p-block elements.
• f-Block Elements (Inner Transition Elements): Placed at the bottom of the periodic table, these elements have their outermost electron(s) in the f subshell. They are also known as lanthanoids (4f series) and actinoids (5f series).

2. d-Block Elements:

• Electronic Configuration: The general electronic configuration is (n-1)d¹⁻¹⁰ ns⁰⁻². The filling of the (n-1)d orbitals occurs after the ns orbitals are filled.
  
• Occurrence: Transition metals are found in various minerals and ores.
  
• Properties:
  
  • Metallic Character: They are all metals, generally hard and high melting and boiling points.
  • Variable Oxidation States: They exhibit multiple oxidation states due to the involvement of (n-1)d electrons in bonding.
  • Formation of Colored Compounds: Many transition metal compounds are colored due to d-d transitions.
  • Catalytic Activity: Many transition metals and their compounds act as catalysts due to their ability to exhibit variable oxidation states and provide a suitable surface for reactions.
  • Formation of Complex Compounds: They readily form complex compounds due to their small size, high charge density, and availability of d orbitals for bonding.
  • Magnetic Properties: Many transition metals and their compounds are paramagnetic due to the presence of unpaired electrons. Some are ferromagnetic.
• Trends in Properties: The chapter discusses trends in atomic radii, ionization enthalpy, electronegativity, and standard electrode potentials across the transition series.
  
• Some Important Compounds: The preparation, properties, and uses of some important compounds of transition metals, such as potassium dichromate (K₂Cr₂O₇) and potassium permanganate (KMnO₄), are discussed.
  

3. f-Block Elements:

• Lanthanoids:
  
  • Electronic Configuration: 4f¹⁻¹⁴ 5d⁰⁻¹ 6s²
  • Lanthanoid Contraction: The gradual decrease in atomic and ionic radii across the lanthanoid series due to the poor shielding effect of 4f electrons. This contraction has several consequences on their properties.
  • Oxidation States: The most common oxidation state is +3.
  • Chemical Reactivity: Lanthanoids are highly reactive metals.
• Actinoids:
  
  • Electronic Configuration: 5f¹⁻¹⁴ 6d⁰⁻² 7s²
  • Oxidation States: They exhibit a wider range of oxidation states than lanthanoids.
  • Radioactivity: All actinoids are radioactive.
  • Chemical Reactivity: Actinoids are highly reactive metals.

Key Differences Between Lanthanoids and Actinoids:

The chapter highlights the key differences in electronic configuration, oxidation states, and radioactivity between lanthanoids and actinoids.

Uses of d- and f-Block Elements:

The chapter concludes with a discussion of the various applications of these elements and their compounds in industry, technology, and everyday life. Many are crucial in catalysis, alloys, pigments, and various specialized applications.

Exercise

1.  Write down the electronic configuration of

 (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Ans : 

(i) Cr³⁺ (Chromium(III) ion)

• Chromium (Cr) has an atomic number of 24. Its neutral electronic configuration is [Ar] 3d⁵ 4s¹.
• Cr³⁺ loses 3 electrons: one from the 4s orbital and two from the 3d orbitals.
• Electronic configuration of Cr³⁺: [Ar] 3d³

(ii) Pm³⁺ (Promethium(III) ion)

• Promethium (Pm) has an atomic number of 61. Its neutral electronic configuration is [Xe] 4f⁵ 6s².
• Pm³⁺ loses 3 electrons: two from the 6s orbital and one from the 4f orbital.
• Electronic configuration of Pm³⁺: [Xe] 4f⁴

(iii) Cu⁺ (Copper(I) ion)

• Copper (Cu) has an atomic number of 29. Its neutral electronic configuration is [Ar] 3d¹⁰ 4s¹.
• Cu⁺ loses 1 electron from the 4s orbital.
• Electronic configuration of Cu⁺: [Ar] 3d¹⁰

(iv) Ce⁴⁺ (Cerium(IV) ion)

• Cerium (Ce) has an atomic number of 58. Its neutral electronic configuration is [Xe] 4f¹ 5d¹ 6s².
• Ce⁴⁺ loses 4 electrons: two from the 6s orbital, one from the 5d orbital, and one from the 4f orbital.
• Electronic configuration of Ce⁴⁺: [Xe]

(v) Co²⁺ (Cobalt(II) ion)

• Cobalt (Co) has an atomic number of 27. Its neutral electronic configuration is [Ar] 3d⁷ 4s².
• Co²⁺ loses 2 electrons from the 4s orbital.
• Electronic configuration of Co²⁺: [Ar] 3d⁵

(vi) Lu²⁺ (Lutetium(II) ion)

• Lutetium (Lu) has an atomic number of 71. Its neutral electronic configuration is [Xe] 4f¹⁴ 5d¹ 6s².
• Lu²⁺ loses 2 electrons: two from the 6s orbital.
• Electronic configuration of Lu²⁺: [Xe] 4f¹⁴ 5d¹

(vii) Mn²⁺ (Manganese(II) ion)

• Manganese (Mn) has an atomic number of 25. Its neutral electronic configuration is [Ar] 3d⁵ 4s².
• Mn²⁺ loses 2 electrons from the 4s orbital.
• Electronic configuration of Mn²⁺: [Ar] 3d⁵

(viii) Th⁴⁺ (Thorium(IV) ion)

• Thorium (Th) has an atomic number of 90. Its neutral electronic configuration is [Rn] 6d² 7s².
• Th⁴⁺ loses 4 electrons: two from the 7s orbital and two from the 6d orbital.
• Electronic configuration of Th⁴⁺: [Rn]

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

Ans : Mn²⁺ compounds are more stable than Fe²⁺ compounds towards oxidation to their +3 state due to the electronic configurations involved.

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 : 

The increasing stability of the +2 oxidation state in the first half of the 3d transition series (Sc to Mn) is due to the increasing energy required to remove the d electrons. As we move across the period, the effective nuclear charge increases, making it progressively harder to ionize the metal atom beyond the +2 state. ¹ Essentially, the higher nuclear charge pulls the d electrons in more strongly, making them less available for further oxidation to higher states. ² This culminates in the stable d⁵ configuration of Mn²⁺, after which the +2 state becomes less stable as electron-electron repulsion in the d orbitals starts to become a factor.

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 : The table shows the rate constant (k) for a reaction at different temperatures (T). To find the activation energy (Ea) and pre-exponential factor (A), plot ln(k) (or log(k)) versus 1/T. The slope of the line will be -Ea/R (or -Ea/2.303R), allowing you to calculate Ea. The intercept will be ln(A) (or log(A)), from which you can find A. Use the Arrhenius equation (k = Ae^(-Ea/RT)) to predict k at other temperatures.

5.  What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Ans : 

(a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

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 : 

CrO₄²⁻ (Chromate ion): Chromium (Cr) is in group 6, and its oxidation state in CrO₄²⁻ is +6.

MnO₄⁻ (Permanganate ion): Manganese (Mn) is in group 7, and its oxidation state in MnO₄⁻ is +7.

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

Ans : 

Lanthanoid Contraction:

Lanthanoid contraction refers to the gradual decrease in the atomic and ionic radii of the lanthanoid elements (elements with atomic numbers 57 to 71) as the atomic number increases.Despite the increase in atomic number (and thus the number of protons and electrons), the size of the atoms and their ions doesn’t increase as expected; instead, it shrinks slightly

Consequences of Lanthanoid Contraction:

Lanthanoid contraction has several important consequences:

1. Similarity in Properties of Second and Third Transition Series: The lanthanoid contraction leads to similar atomic radii for the elements of the second (4d) and third (5d) transition series. For example, Zirconium (Zr) and Hafnium (Hf), which are in the same group, have very similar radii due to lanthanoid contraction. This similarity in size leads to similar chemical properties, making it difficult to separate these elements.  
2. Basicity of Oxides: The decrease in ionic radii of M³⁺ ions across the lanthanoid series leads to an increase in the covalent character of M-O bonds in their oxides. As a result, the basicity of the oxides of lanthanoids decreases with increasing atomic number. The earlier lanthanoids are more basic than the later ones.  
3. Difficulty in Separation of Lanthanoids: Because the lanthanoids have very similar chemical properties due to their nearly identical ionic radii, their separation becomes challenging. Separation methods often rely on very subtle differences in their properties, such as ion-exchange chromatography.  
4. Trends in Ionization Enthalpies: The lanthanoid contraction influences the ionization enthalpies of the transition elements. The ionization enthalpies of the 5d series are higher than those of the 4d series due to the increased effective nuclear charge.  
5. Hydration of Ions: The degree of hydration of lanthanoid ions is also affected. Smaller ions have a greater charge density and thus are more heavily hydrated. This affects their mobility and separation.

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 Transition Elements:

• Incomplete d subshell: Transition elements have an incomplete (n-1)d subshell in their atomic or ionic states. This is the defining characteristic.  
• Metallic Properties: They are all metals, possessing properties like luster, hardness, high melting and boiling points, and good conductivity of heat and electricity.  
• Variable Oxidation States: Transition elements exhibit a range of oxidation states due to the involvement of d electrons in bonding.  
• Formation of Colored Compounds: Many transition metal compounds are colored because of the d-d transitions of electrons within their ions.  
• Catalytic Activity: Many transition metals and their compounds act as catalysts in chemical reactions.  
• Formation of Complex Compounds: They have a strong tendency to form coordination complexes due to their small size, high charge density, and availability of d orbitals for bonding.
• Magnetic Properties: Many transition metals and their compounds are paramagnetic (attracted to magnetic fields) due to unpaired electrons in their d orbitals. Some are ferromagnetic (strongly magnetic).  

Why are they called Transition Elements?

They are called “transition” elements because their properties represent a transition between the highly reactive s-block elements (like alkali and alkaline earth metals) and the p-block elements (which include nonmetals and metalloids). They bridge the gap in properties between these two blocks of the periodic table.  

Which d-block elements may not be considered transition elements?

The elements in Group 12 (Zinc, Cadmium, Mercury, and Copernicium) are often debated. They have a completely filled d¹⁰ configuration in their ground state. Some chemists argue that since they don’t have an incomplete d subshell in their elemental form, they shouldn’t be strictly classified as transition elements. However, they do form ions with incomplete d subshells (like Hg²⁺), and they do exhibit some properties typical of transition metals. Therefore, they are usually included in the d-block, even if their classification as true “transition elements” is sometimes debated.

9. In what way are the electronic configuration of the transition elements different from non-transition elements?

Ans : 

Transition Elements: These elements have their outermost electrons in the d subshell (and sometimes the s subshell as well). Specifically, the (n-1)d subshell is being filled. ¹ This means that the d orbitals are involved in bonding and are responsible for many of the characteristic properties of transition metals (variable oxidation states, formation of colored compounds, catalytic activity, etc.).

Non-transition Elements: These elements have their outermost electrons in the s and p subshells. The d subshell is either completely filled or empty. Since the d orbitals are not involved in bonding to the same extent, non-transition elements generally exhibit fewer oxidation states and do not show the same range of catalytic and complex-forming abilities as transition metals.

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

Ans : While +3 is the dominant and most stable oxidation state for most lanthanoids, +2 and +4 states can occur, primarily when they lead to particularly stable f-electron configurations.

11. 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) 

• Reason: Paramagnetism arises from the presence of unpaired electrons. These unpaired electrons spin, creating a magnetic moment. When placed in an external magnetic field, these spins align, resulting in paramagnetism. The more unpaired electrons, the stronger the paramagnetic behavior.

(ii) 

• Reason: Enthalpy of atomization is the energy required to break the metallic bonds and convert the metal into individual gaseous atoms. Transition metals have strong metallic bonds due to the involvement of d electrons in addition to s electrons in the metallic bonding. The greater the number of unpaired d electrons, the stronger the interatomic interactions and hence, the higher the enthalpy of atomization.

(iii) 

• Reason: The color of transition metal compounds is due to d-d transitions. When ligands approach a transition metal ion, the d orbitals split into different energy levels. Electrons can jump between these d orbitals by absorbing specific wavelengths of light in the visible region. The transmitted light, which is complementary to the absorbed light, gives the compound its color. The specific color depends on the metal ion, the ligands surrounding it, and the geometry of the complex.

(iv) 

• Reason: Transition metals and their compounds make excellent catalysts due to a combination of factors:
  • Variable Oxidation States: They can readily change their oxidation states, allowing them to participate in redox reactions and facilitate electron transfer.
  • Surface Area: Transition metals provide a large surface area for reactant molecules to adsorb onto, increasing the chances of effective collisions.
  • Complex Formation: They can form temporary bonds with reactant molecules, forming intermediate complexes. This brings reactants closer together in the correct orientation, weakening existing bonds and facilitating the formation of new ones. This lowers the activation energy of the reaction.
  • Availability of d-orbitals: The presence of empty or partially filled d-orbitals allows for the accommodation of reactant molecules and the formation of activated complexes.

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

Ans : 

Interstitial compounds are formed when small atoms, such as hydrogen, carbon, nitrogen, or boron, are trapped in the spaces (interstices) between the atoms in the crystal lattice of a metal.  

Why are they well known for transition metals?

Transition metals have a unique combination of characteristics that make them prone to forming interstitial compounds:  

• Vacant spaces in the crystal lattice: Transition metals tend to form crystal lattices with “holes” or “voids” between the metal atoms. These spaces are just the right size to accommodate small nonmetal atoms.  
• Variable oxidation states: Transition metals can exhibit multiple oxidation states, allowing them to form compounds with a variety of stoichiometries, including those where the small atoms occupy interstitial sites.  
• Atomic size: Transition metal atoms are generally larger than the nonmetal atoms that form interstitial compounds. This size difference allows the smaller atoms to fit comfortably within the metal lattice without significantly disrupting it.

13.  How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Ans : 

Transition Metals:

• Multiple Oxidation States: Transition metals exhibit a wider range of oxidation states. This is because both the ns and (n-1)d electrons can participate in bonding. The energy difference between these orbitals is small, allowing for the loss of varying numbers of electrons.  
• Oxidation States Differ by 1: The oxidation states of transition metals typically differ by 1 (e.g., Fe²⁺ and Fe³⁺).  
• Examples:
  • Iron (Fe): Exhibits +2, +3, +4, +6 oxidation states.
  • Manganese (Mn): Exhibits +2, +3, +4, +5, +6, +7 oxidation states.
  • Vanadium (V): Exhibits +2, +3, +4, +5 oxidation states.

Non-Transition Metals:

• Oxidation States Differ by 2: The oxidation states of non-transition metals often differ by 2 (e.g., Sn²⁺ and Sn⁴⁺). This is related to the inert pair effect, where the ns electrons become less available for bonding in heavier elements.  
• Examples:
  • Carbon (C): Exhibits -4, -2, +2, +4 oxidation states.
  • Nitrogen (N): Exhibits -3, -2, -1, +1, +2, +3, +4, +5 oxidation states.
  • Sulfur (S): Exhibits -2, +2, +4, +6 oxidation states.

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 : 

Preparation of Potassium Dichromate (K₂Cr₂O₇) from Iron Chromite Ore (FeCr₂O₄):

The process involves several steps:

1. Roasting with Sodium Carbonate:
   
   • The chromite ore is finely ground and mixed with sodium carbonate (Na₂CO₃) and calcium oxide (CaO) (or sometimes just sodium carbonate).
   • This mixture is roasted in a reverberatory furnace in the presence of excess air at high temperatures (around 1000-1200°C).
   • The following reaction occurs: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂  
2. Conversion to Sodium Dichromate:
   
   • The roasted mass contains sodium chromate (Na₂CrO₄).
   • This is leached with water to dissolve the sodium chromate, leaving behind insoluble iron(III) oxide (Fe₂O₃) and other impurities.
   • The filtered solution containing sodium chromate is then treated with concentrated sulfuric acid (H₂SO₄).  
   • This converts the sodium chromate to sodium dichromate (Na₂Cr₂O₇): 2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O  
3. Conversion to Potassium Dichromate:
   
   • The sodium dichromate solution is evaporated to obtain solid Na₂Cr₂O₇.
   • This solid is then dissolved in water and treated with potassium chloride (KCl).
   • Because potassium dichromate (K₂Cr₂O₇) is less soluble than sodium chloride (NaCl), it crystallizes out as orange crystals upon cooling: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl  

Effect of Increasing pH on Potassium Dichromate Solution:

Potassium dichromate exists in equilibrium with potassium chromate in aqueous solution:  

Cr₂O₇²⁻ (dichromate, orange) + H₂O ⇌ 2CrO₄²⁻ (chromate, yellow) + 2H⁺

• Acidic Conditions (low pH): The equilibrium shifts to the left, favoring the formation of the dichromate ion (Cr₂O₇²⁻). The solution appears orange.  
• Basic Conditions (high pH): The equilibrium shifts to the right, favoring the formation of the chromate ion (CrO₄²⁻). The solution turns yellow.

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 : 

(i) Iodide (I⁻):

Potassium dichromate oxidizes iodide ions (I⁻) to iodine (I₂).  

• Ionic Equation: Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂

(ii) Iron(II) Solution (Fe²⁺):

Potassium dichromate oxidizes iron(II) ions (Fe²⁺) to iron(III) ions (Fe³⁺).  

• Ionic Equation: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺  

(iii) Hydrogen Sulfide (H₂S):

Potassium dichromate oxidizes hydrogen sulfide (H₂S) to sulfur (S).

• Ionic Equation: Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 7H₂O + 3S

16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (II) ions (ii) S02 and (iii) oxalic acid? Write the ionic, equations for the reactions.

Ans : 

Potassium permanganate is prepared from pyrolusite ore (MnO₂). The process involves several steps:  

1. Fusion with Alkali:
   
   • The pyrolusite ore is finely ground and mixed with potassium hydroxide (KOH) and an oxidizing agent like potassium nitrate (KNO₃) or potassium chlorate (KClO₃).  
   • This mixture is fused in a furnace at high temperatures.
   • The following reaction occurs: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O (Potassium manganate, green)
2. Electrolytic Oxidation:
   
   • The fused mass is dissolved in water, and the green solution of potassium manganate (K₂MnO₄) is subjected to electrolytic oxidation.
   • In this process, manganate ions (MnO₄²⁻) are oxidized to permanganate ions (MnO₄⁻): MnO₄²⁻ → MnO₄⁻ + e⁻
3. Crystallization:
   
   • The resulting solution containing permanganate ions is evaporated to concentrate it.
   • Upon cooling, dark purple crystals of potassium permanganate (KMnO₄) separate out.

Reactions of Acidified Permanganate Solution

Acidified permanganate solution is a powerful oxidizing agent. The permanganate ion (MnO₄⁻) is reduced to Mn²⁺ in acidic medium:  

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Reactions with:

(i) Iron(II) Ions (Fe²⁺):

Acidified permanganate oxidizes iron(II) ions (Fe²⁺) to iron(III) ions (Fe³⁺).  

• Ionic Equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

(ii) Sulfur Dioxide (SO₂):

Acidified permanganate reduces sulfur dioxide (SO₂) to sulfate ions (SO₄²⁻).

• Ionic Equation: 2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺

(iii) Oxalic Acid (H₂C₂O₄):

Acidified permanganate oxidizes oxalic acid (H₂C₂O₄) to carbon dioxide (CO₂).  

• Ionic Equation: 2MnO₄⁻ + 16H⁺ + 5H₂C₂O₄ → 2Mn²⁺ + 10CO₂ + 8H₂O

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

Cr2+/Cr   –> -0.9 V

Mn2+/Mn  –> -1.2V

Fe2+/Fe     –> -0.4 V

Cr3+/Cr2+  –> -0.4 V

Mn3+/Mn2+   –>+ 1.5V

Fe3+/Fe2+   –>+ 0.8V

Use this data tp 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) Stability of M³⁺ ions in acid solution:

The stability of the +3 oxidation state compared to the +2 state is determined by the standard potential (E°) for the M³⁺/M²⁺ couple. A higher E° value indicates a greater tendency for the M³⁺ ion to be reduced to M²⁺, meaning the M³⁺ state is less stable and readily gets reduced. Conversely, a lower E° implies a greater stability of the M³⁺ state.

• Cr³⁺/Cr²⁺: E° = -0.4 V. This relatively low value indicates that Cr³⁺ has a greater tendency to remain in the +3 state. It’s more stable than Mn³⁺ or Fe³⁺.
  
• Mn³⁺/Mn²⁺: E° = +1.5 V. This high positive value indicates that Mn³⁺ readily gets reduced to Mn²⁺. Mn³⁺ is therefore the least stable among the three ions in the +3 state.
  
• Fe³⁺/Fe²⁺: E° = +0.8 V. This value is intermediate between Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺. Fe³⁺ is less stable than Cr³⁺ but more stable than Mn³⁺.
  

In summary: Cr³⁺ is the most stable in the +3 state, followed by Fe³⁺, and then Mn³⁺ (which is the least stable).

(ii) Ease of oxidation of M to M²⁺:

The ease of oxidation of M to M²⁺ is related to the standard potential (E°) for the M²⁺/M couple. A lower (more negative) E° value indicates a greater ease of oxidation.

• Cr²⁺/Cr: E° = -0.9 V. This relatively low value indicates that Chromium metal is readily oxidized to Cr²⁺.
  
• Mn²⁺/Mn: E° = -1.2 V. This is the most negative value, meaning Manganese metal is the easiest to oxidize to Mn²⁺.
  
• Fe²⁺/Fe: E° = -0.4 V. Iron is oxidized more easily than chromium (since both are -0.4V), but less easily than manganese.
  

In summary: Manganese is the easiest to oxidize, followed by iron and then chromium.

18. Predict which of the following will be coloured in aqueous solution?

Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+, Co2+.

Ans : 

• Ti³⁺: Titanium (Ti) is [Ar] 3d² 4s². Ti³⁺ loses 3 electrons to become [Ar] 3d¹. (Partially filled d orbital) – Colored
  
• V³⁺: Vanadium (V) is [Ar] 3d³ 4s². V³⁺ loses 3 electrons to become [Ar] 3d². (Partially filled d orbital) – Colored
  
• Cu⁺: Copper (Cu) is [Ar] 3d¹⁰ 4s¹. Cu⁺ loses 1 electron to become [Ar] 3d¹⁰. (Completely filled d orbital) – Colorless
  
• Sc³⁺: Scandium (Sc) is [Ar] 3d¹ 4s². Sc³⁺ loses 3 electrons to become [Ar]. (Empty d orbital) – Colorless
  
• Mn²⁺: Manganese (Mn) is [Ar] 3d⁵ 4s². Mn²⁺ loses 2 electrons to become [Ar] 3d⁵. (Partially filled d orbital) – Colored
  
• Fe³⁺: Iron (Fe) is [Ar] 3d⁶ 4s². Fe³⁺ loses 3 electrons to become [Ar] 3d⁵. (Partially filled d orbital) – Colored
  
• Co²⁺: Cobalt (Co) is [Ar] 3d⁷ 4s². Co²⁺ loses 2 electrons to become [Ar] 3d⁵. (Partially filled d orbital) – Colored

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

Ans : The stability of the +2 oxidation state in the first transition series is a result of the interplay between increasing ionization energy, the stability associated with half-filled and fully filled d subshells, and hydration enthalpy. This leads to a general trend of increasing stability of the +2 state from Sc to Zn.

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

(i)electronic configuration,

(ii)atomic and ionic sizes and

(iii)oxidation state

(iv)chemical reactivity.

Ans : 

(i) Electronic Configuration:

• Lanthanoids: The general electronic configuration is [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s². Electrons are primarily filled in the 4f subshell. There are some exceptions, like Gadolinium (Gd), which has a 5d¹ configuration to achieve a half-filled 4f subshell.
  
• Actinoids: The general electronic configuration is [Rn] 5f¹⁻¹⁴ 6d⁰⁻² 7s². Electrons are primarily filled in the 5f subshell, although there are more irregularities than in the lanthanoids due to the comparable energies of the 5f, 6d, and 7s orbitals. Many actinoids have electrons in the 6d subshell.

Key Difference: Actinoids show greater variability in filling their f and d orbitals than lanthanoids.

(ii) Atomic and Ionic Sizes:

• Lanthanoids: Exhibit lanthanoid contraction – a gradual decrease in atomic and ionic radii with increasing atomic number. This is due to the poor shielding effect of 4f electrons, leading to increased effective nuclear charge.  
• Actinoids: Also exhibit actinoid contraction, which is similar to lanthanoid contraction but generally more pronounced. Again, this is due to the poor shielding by 5f electrons.

Key Difference: Actinoid contraction is greater than lanthanoid contraction.  

(iii) Oxidation States:

• Lanthanoids: The most common and stable oxidation state is +3. Some lanthanoids show +2 and +4 states, but these are less common and usually occur when they lead to a noble gas configuration (f⁰), a half-filled f subshell (f⁷), or a filled f subshell (f¹⁴).  
• Actinoids: Show a wider range of oxidation states, from +2 to +7, with +3 being the most common. The greater involvement of 5f, and particularly 6d, electrons allows for this greater variability.

Key Difference: Actinoids exhibit a wider range of oxidation states than lanthanoids.  

(iv) Chemical Reactivity:

• Lanthanoids: Generally highly reactive metals, similar to alkaline earth metals. They readily react with oxygen, water, acids, and halogens. Reactivity decreases somewhat across the series as the ionization energy increases.
  
• Actinoids: Also highly reactive metals. Due to their radioactivity, special handling precautions are required. Their chemical reactivity is complex and varies depending on the specific element and reaction conditions.  

Key Difference: Both are reactive, but the radioactivity of actinoids adds a layer of complexity to their chemical reactivity.

Also, some of the more well-characterized reactions of actinoids (e.g., those involving uranium and plutonium) are of great practical importance.

21.  How would you account for the following:

(i) Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidizing.

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

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

Ans : 

(i) Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidizing.

• Cr²⁺ (d⁴): Cr²⁺ readily loses an electron to become Cr³⁺ (d³). The d³ configuration is relatively stable (though not as stable as d⁵) due to the absence of electron-electron repulsions that would be present in configurations with more electrons. Since it readily loses an electron, it acts as a strong reducing agent.
  
• Mn³⁺ (d⁴): Mn³⁺ readily gains an electron to become Mn²⁺ (d⁵). The d⁵ configuration is a particularly stable configuration due to the half-filled d subshell, which minimizes electron-electron repulsion and maximizes exchange energy. Since it readily gains an electron, it acts as a strong oxidizing agent.

(ii) Co²⁺ is stable in aqueous solution but is easily oxidized in the presence of complexing reagents.

• Co²⁺ (d⁷): In aqueous solution, Co²⁺ exists as a hydrated ion, [Co(H₂O)₆]²⁺. This hydrated complex is relatively stable.
  
• Oxidation by Complexing Reagents: When complexing reagents are added, they form more stable complexes with Co³⁺ (d⁶) than with Co²⁺. This is due to the higher charge density of Co³⁺, which allows it to form stronger bonds with the ligands. The formation of the more stable Co³⁺ complex lowers the energy of the Co³⁺ state, making the oxidation of Co²⁺ to Co³⁺ thermodynamically favorable. In essence, the complexing agent removes Co³⁺ from the equilibrium, driving the oxidation forward.

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

• Reason: A d¹ configuration means that the single electron in the d subshell experiences significant electron-electron repulsion if it were to pair up with another electron in the same orbital. This repulsion makes the d¹ configuration relatively high in energy and therefore unstable. Ions with a d¹ configuration will readily lose this electron to achieve a noble gas configuration or gain an electron to achieve a more stable configuration (like d² or higher). This instability is reflected in the fact that ions with d¹ configurations are strong reducing agents.

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

Ans : Disproportionation is a type of redox reaction where a single chemical species (an atom, ion, or molecule) is simultaneously both oxidized and reduced, resulting in the formation of two distinct products. In simpler terms, it’s when one substance in an intermediate oxidation state converts into two products, one in a higher oxidation state and the other in a lower oxidation state.  

Examples of Disproportionation Reactions in Aqueous Solution:

1. Disproportionation of Copper(I) ion (Cu⁺):

Copper(I) ion (Cu⁺) is unstable in aqueous solution and undergoes disproportionation to form copper(II) ions (Cu²⁺) and copper metal (Cu):  

2 Cu⁺ (aq) → Cu²⁺ (aq) + Cu (s)

• In Cu⁺, copper has an oxidation state of +1.
• In Cu²⁺, copper has an oxidation state of +2 (oxidation).  
• In Cu, copper has an oxidation state of 0 (reduction).

2. Disproportionation of Chlorine in Basic Solution:

When chlorine gas (Cl₂) reacts with a cold, dilute solution of a base like sodium hydroxide (NaOH), it undergoes disproportionation to form chloride ions (Cl⁻) and hypochlorite ions (ClO⁻):  

Cl₂ (g) + 2 NaOH (aq) → NaCl (aq) + NaClO (aq) + H₂O (l)

• In Cl₂, chlorine has an oxidation state of 0.  
• In Cl⁻, chlorine has an oxidation state of -1 (reduction).
• In ClO⁻, chlorine has an oxidation state of +1 (oxidation).

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

Ans : Copper (Cu) exhibits the +1 oxidation state most frequently among the first transition series metals.  

Copper’s electronic configuration is [Ar] 3d¹⁰4s¹. While it can lose two electrons to form the more common Cu²⁺ (+2) state, the +1 state (Cu⁺) is also readily accessible and relatively stable.

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⁺:

• Mn is [Ar] 3d⁵ 4s²
• Mn³⁺ loses 3 electrons (2 from 4s, 1 from 3d) to become [Ar] 3d⁴  
• Unpaired electrons: 4

Cr³⁺:

• Cr is [Ar] 3d⁵ 4s¹
• Cr³⁺ loses 3 electrons (1 from 4s, 2 from 3d) to become [Ar] 3d³
• Unpaired electrons: 3

V³⁺:

• V is [Ar] 3d³ 4s²
• V³⁺ loses 3 electrons (2 from 4s, 1 from 3d) to become [Ar] 3d²
• Unpaired electrons: 2

Ti³⁺:

• Ti is [Ar] 3d² 4s²
• Ti³⁺ loses 3 electrons (2 from 4s, 1 from 3d) to become [Ar] 3d¹
• Unpaired electrons: 1

Therefore, Cr³⁺ is the most stable in aqueous solution among the ions listed due to its d³ configuration.

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.

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

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

Ans : 

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

• Reason: The oxidation state of the metal influences the nature of the bond formed with oxygen. In lower oxidation states, the metal ion has a lower charge density and interacts less strongly with oxygen. The oxide tends to be more ionic, and the oxygen atoms can readily accept protons (H⁺), leading to basic character. As the oxidation state increases, the metal ion has a higher charge density and pulls the electrons of the oxygen atoms more strongly towards itself. This leads to a more covalent character in the metal-oxygen bond. In the highest oxidation states, the metal atom is so electron-deficient that it pulls electron density away from the oxygen, making the oxide more acidic.
  
• Examples:
  
  • MnO: Manganese in its +2 oxidation state forms MnO, which is a basic oxide.
  • Mn₂O₇: Manganese in its +7 oxidation state forms Mn₂O₇, which is a strongly acidic oxide.
  • CrO: Chromium in its +2 oxidation state forms CrO, which is a basic oxide.
  • CrO₃: Chromium in its +6 oxidation state forms CrO₃, which is an acidic oxide.
  • V₂O₃: Vanadium in its +3 oxidation state forms V₂O₃, which is a basic oxide.
  • V₂O₅: Vanadium in its +5 oxidation state forms V₂O₅, which is an amphoteric oxide.

(ii) 

• Reason: Both oxygen and fluorine are highly electronegative elements. They have a strong tendency to attract electrons. This allows them to stabilize the transition metal in its highest oxidation state. Fluorine, being the most electronegative element, is particularly effective at this. Oxides also stabilize high oxidation states due to the strong metal-oxygen bonds formed.
  
• Examples:
  
  • Osmium (Os): Osmium reaches its highest known oxidation state of +8 in OsO₄ (oxide).
  • Ruthenium (Ru): Ruthenium also reaches its highest known oxidation state of +8 in RuO₄ (oxide).
  • Chromium (Cr): Chromium reaches its highest oxidation state of +6 in CrO₃ (oxide) and CrF₆ (fluoride).

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

• Reason: In oxoanions, the metal is bonded to several oxygen atoms. The oxygen atoms, being highly electronegative, draw electron density away from the metal. This allows the metal to achieve a high oxidation state. Also, the structure of the oxoanion can help to stabilize the high oxidation state.
  
• Examples:
  
  • Manganese (Mn): Exhibits its highest oxidation state of +7 in permanganate ion, MnO₄⁻.
  • Chromium (Cr): Exhibits its highest oxidation state of +6 in chromate ion, CrO₄²⁻, and dichromate ion, Cr₂O₇²⁻.
  • Vanadium (V): Exhibits its highest oxidation state of +5 in vanadate ion, VO₄³⁻.

26. Indicate the steps in the preparation of:

(i)K2Cr207from chromite ore

(ii)KMn04 from pyrolusite ore.

Ans : 

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

Ans : 

What are Alloys?

Alloys are homogeneous mixtures (solid solutions) of two or more metals, or a metal and a nonmetal. They are created to enhance certain desirable properties of the base metal or to combine the properties of different metals. Alloys often have different physical and chemical properties compared to their constituent elements.  

Mischmetal: An Alloy Containing Lanthanoid Metals  

One important alloy that contains lanthanoid metals is mischmetal.  

• Composition: Mischmetal is an alloy of approximately 95% lanthanoid metals and 5% iron, along with trace amounts of other elements like sulfur, carbon, calcium, and aluminum. The lanthanoid metals present are primarily cerium and lanthanum, with smaller amounts of other lanthanoids.  
• Uses: Mischmetal has a variety of applications:
  
  • Lighter Flints: Mischmetal is pyrophoric (it sparks when scratched), making it ideal for use in lighter flints.  
  • Magnesium Alloys: It is added to magnesium-based alloys to increase their strength and hardness, making them suitable for use in aircraft components and other demanding applications.  
  • Deoxidizer: Mischmetal is used as a deoxidizer in the production of steel and other metals.  
  • Alloying Agent: It is used as an alloying agent in various metallurgical applications to improve the properties of other metals.

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 :
Lanthanides: Atomic numbers 57-71  

Actinides: Atomic numbers 89-103  

Checking the Given Atomic Numbers

Let’s examine the provided atomic numbers:

• 29: Copper (Cu) – Not an inner transition element
• 59: Praseodymium (Pr) – Inner transition element (Lanthanide)  
• 74: Tungsten (W) – Not an inner transition element
• 95: Americium (Am) – Inner transition element (Actinide)  
• 102: Nobelium (No) – Inner transition element (Actinide)  
• 104: Rutherfordium (Rf) – Not an inner transition element

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

Ans : Lanthanides exhibit a limited number of oxidation states, primarily +2, +3, and +4, with +3 being the most common. This is due to the significant energy gap between the 4f, 5d, and 6s subshells. Similarly, the dominant oxidation state of actinides is also +3, but they display a wider range of oxidation states. For instance, uranium (Z = 92) and plutonium (Z = 94) can exist in +3, +4, +5, and +6 oxidation states, while neptunium (Z = 93) exhibits +3, +4, +5, and +7 states. This variability arises from the small energy difference between the 5f, 6d, and 7s orbitals in actinides.

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

Ans : The last element in the actinoid series is Lawrencium (Lr), atomic number 103.  

Electronic Configuration:

The electronic configuration of Lawrencium is [Rn] 5f¹⁴ 6d¹ 7s².

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 : 

32. Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to co-relate this type of behaviour with the electronic configuration of these elements.

Ans : 

The +4 oxidation state is exhibited by elements such as cerium (58Ce), praseodymium (59Pr), and terbium (65Tb).

The +2 oxidation state is observed in neodymium (60Nd), samarium (62Sm), europium (63Eu), thulium (69Tm), and ytterbium (70Yb).

In general, the +2 oxidation state is shown by elements with the electron configuration 5d⁰6s², allowing the easy loss of two electrons. Similarly, the +4 oxidation state is exhibited by elements that, upon losing four electrons, attain a configuration close to either 4f⁰ or 4f⁷.

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

(i)Electronic configuration

(ii)Oxidation states

(iii)Chemical reactivity

Ans : 

(i) Electronic Configuration:

• Lanthanoids: Primarily fill the 4f subshell. Their general configuration is [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s². The 5d orbital may or may not have an electron.
• Actinoids: Primarily fill the 5f subshell. Their general configuration is [Rn] 5f¹⁻¹⁴ 6d⁰⁻² 7s². Actinoids show more variability in the occupancy of their 5f and 6d orbitals due to smaller energy differences between them.

Key Difference: Actinoids have a greater tendency for electrons to occupy the 6d orbitals compared to the lanthanoids, where 5d occupation is less common.

(ii) Oxidation States:

• Lanthanoids: Primarily exhibit a +3 oxidation state. Some show +2 and +4, but these are less common and usually tied to achieving stable f-subshell configurations (f⁰, f⁷, f¹⁴).
• Actinoids: Exhibit a wider range of oxidation states, from +2 to +7, with +3 being the most common. The smaller energy difference between 5f, 6d, and 7s levels allows for more varied oxidation states.  

Key Difference: Actinoids show a greater diversity of oxidation states compared to lanthanoids.  

(iii) Chemical Reactivity:

• Lanthanoids: Highly reactive metals, similar to alkaline earth metals. They readily react with oxygen, water, acids, and halogens. They form ionic compounds predominantly.  
• Actinoids: Also highly reactive metals, but their reactivity is complicated by their radioactivity. They react with many of the same substances lanthanoids do. Actinoids can display more covalent character in their bonding compared to lanthanoids.

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

Ans : 

Atomic Number 61 (Promethium, Pm):

• [Xe] 4f⁵ 6s²  

Atomic Number 91 (Protactinium, Pa):

• [Rn] 5f² 6d¹ 7s²  

Atomic Number 101 (Mendelevium, Md):

• [Rn] 5f¹¹ 7s²

Atomic Number 109 (Meitnerium, Mt):

• [Rn] 5f¹⁴ 6d⁷ 7s²

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

(ii)oxidation states

(iii)ionisation enthalpies and

(iv)atomic sizes

Ans : 

Property	First Series (3d)	Second Series (4d)	Third Series (5d)
Electronic Config.	(n-1)d¹⁻¹⁰ ns¹⁻²	(n-1)d¹⁻¹⁰ ns¹⁻²	(n-1)d¹⁻¹⁰ ns¹⁻²
Oxidation States	+2, +3 common, highest up to +7	Wider range, higher states more common	Wider range, higher states more common
Ionization Enthalpy	Increases across period	Higher than 3d series	Comparable to or slightly higher than 4d series
Atomic Radii	Decreases across period	Larger than 3d series	Similar to 4d series due to lanthanide contraction

36. Write down the number of 3d electrons in each of the following ions:Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe2+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Ans : 

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

Ans : 

The heavier transition elements belong to the fourth (4d), fifth (5d), and sixth (6d) transition series. Their properties differ from those of the first (3d) series due to the following reasons:

(i) The atomic radii of elements in the 4d and 5d series are generally larger due to the presence of more electron shells. However, the difference between the atomic radii of 4d and 5d transition elements is relatively small due to the effect of lanthanoid contraction.

(ii) Due to stronger interatomic bonding, the melting and boiling points of the elements in the 4d and 5d series are higher than those of the 3d series.

(iii) Ionization enthalpies are expected to decrease as we move from one transition series to the next. However, the ionization enthalpies of the 5d series elements are higher than those of the other two series due to lanthanoid contraction. This contraction leads to a decrease in atomic size and an increase in effective nuclear charge, resulting in higher ionization energy, particularly in the 3d elements.

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

Ans :