Class 12 Chemistry Chapter 4 The d and f Block Elements ncert solutions

The d and f Block Elements Class 12 ncert solutions: Chemistry Class 12 Chapter 4 exercise solutions

Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 4
Chapter Name	The d and f Block Elements ncert solutions
Category	Ncert Solutions
Medium	English

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Question 4.1: Write down the electronic configuration of (i) Cr³⁺ (ii) Pm³⁺ (iii) Cu⁺ (iv) Ce⁴⁺ (v) Co²⁺ (vi) Lu²⁺ (vii) Mn²⁺  (viii) Th⁴⁺

Solution 4.1:

• (i) Cr³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ Or, [Ar]¹⁸ 3d³
• (ii) Pm³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f⁴ Or, [Xe]⁵⁴ 3d³
• (iii) Cu⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ Or, [Ar]¹⁸ 3d¹⁰
• (iv) Ce⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ Or, [Xe]⁵⁴
• (v) Co²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ Or, [Ar]¹⁸ 3d⁷
• (vi) Lu²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹ Or, [Xe]⁵⁴ 2f¹⁴ 3d³
• (vii) Mn²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ Or, [Ar]¹⁸ 3d⁵
• (viii) Th⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6s⁶ Or, [Rn]⁸⁶

Question 4.2: Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?

Solution 4.2:

Electronic configuration of Mn²⁺ is [Ar]¹⁸ 3d⁵.

Electronic configuration of Fe²⁺ is [Ar]¹⁸ 3d⁶.

Mn²⁺ compounds are more stable than Fe²⁺ towards oxidation to the +3 state because Mn²⁺ has a half-filled 3d⁵ electronic configuration, which is especially stable due to symmetry and exchange energy. In contrast, Fe²⁺ has a 3d⁶ configuration, which is less stable, making Fe²⁺ more easily oxidized to Fe³⁺.

Question 4.3: Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Solution 4.3: In the first half of the first-row transition elements, the +2 oxidation state becomes more stable with increasing atomic number due to the increasing number of 3d electrons. As the atomic number increases from Sc to Mn, the energy required to remove additional electrons from the d-orbitals decreases. This is because the 3d subshell is progressively filling up, leading to greater stabilization of the +2 state, as the d-electrons become more stable and harder to remove. Additionally, the increase in effective nuclear charge helps stabilize the +2 state by attracting the outer electrons more strongly.

Question 4.4: 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.

Solution 4.4: The electronic configurations play a crucial role in determining the stability of oxidation states in the first series of transition elements. This is because the relative stability of different oxidation states depends on the energy and stability of the d-electron configuration after electrons are removed or added.

Influence of Electronic Configurations on Stability of Oxidation States:

1. Half-filled and fully filled configurations:
   Half-filled (d⁵) and fully filled (d¹⁰) d-orbital configurations are particularly stable. For example:

• Mn (Manganese): Mn²⁺ has a 3d⁵ configuration, which is half-filled and highly stable. This makes the +2 state more stable for manganese, and Mn³⁺ is less stable since removing an additional electron would disturb this stable configuration.
• Zn (Zinc): Zn²⁺ has a 3d¹⁰ configuration, which is fully filled and very stable. Hence, the +2 oxidation state is the most stable for zinc, and higher oxidation states are rarely observed.

1. Increasing number of d-electrons:
   As the number of d-electrons increases across the series, the stability of higher oxidation states also tends to increase initially but then decreases due to electron repulsion in the d-orbitals. For example:

• Fe (Iron): Fe²⁺ (3d⁶) is less stable compared to Fe³⁺ (3d⁵), as Fe³⁺ has a half-filled, stable d⁵ configuration.
• Cr (Chromium): Cr³⁺ (3d³) is more stable than Cr²⁺ (3d⁴), as Cr³⁺ represents a more stable configuration due to fewer unpaired electrons and lower electron-electron repulsion.

The stability of oxidation states in the first-row transition elements is largely dictated by electronic configurations, with half-filled (d⁵) and fully filled (d¹⁰) states being particularly stable. As the d-electrons increase across the series, the relative stability of the +2, +3, and higher oxidation states varies accordingly.

Question 4.5: 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?

Solution 4.5:

	Electronic configuration in ground state	Stable oxidation states
(i)	3d3 (Vanadium)	+2, +3, +4 and +5
(ii)	3d5 (Chromium)	+3, +4, +6
(iii)	3d5 (Manganese)	+2, +4, +6, +7
(iv)	3d8(Cobalt)	+2, +3
(v)	3d4	There is no3d4 configuration in ground state

Question 4.6: Name the oxometal anions of the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.

Solution 4.6: In the first series of transition metals, the metals can form oxometal anions in which they exhibit an oxidation state equal to their group number. These oxometal anions are typically formed with oxygen atoms, and the metal achieves its highest oxidation state. Here are the examples where the oxidation state of the metal equals its group number:

1. Vanadate (VO₄³⁻):

• Vanadium (V) is in Group 5.
• In the vanadate ion (VO₄³⁻), vanadium exhibits an oxidation state of +5, which is equal to its group number.

2. Chromate (CrO₄²⁻) and Dichromate (Cr₂O₇²⁻):

• Chromium (Cr) is in Group 6.
• In both the chromate ion (CrO₄²⁻) and the dichromate ion (Cr₂O₇²⁻), chromium exhibits an oxidation state of +6, matching its group number.

3. Permanganate (MnO₄⁻):

• Manganese (Mn) is in Group 7.
• In the permanganate ion (MnO₄⁻), manganese is in the +7 oxidation state, which equals its group number.

These are the common oxometal anions where the metal exhibits an oxidation state equal to its group number in the first transition series.

Question 4.7: What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Solution 4.7: Lanthanoid Contraction:

Lanthanoid contraction refers to the gradual decrease in atomic and ionic radii of the lanthanide series elements (from La to Lu) as the atomic number increases. This occurs despite the addition of more electrons because the increasing nuclear charge is not fully offset by the added electrons in the 4f subshell, which have poor shielding effect. As a result, the nucleus pulls the electrons closer, causing the size of the atoms and ions to contract.

Consequences of Lanthanoid Contraction:

• Influence on chemical reactivity: The decrease in ionic radii leads to a slight increase in ionization energies, affecting the chemical reactivity and basicity of compounds.
• Similarity in size between 4d and 5d transition elements: Elements like Zr (4d) and Hf (5d) have nearly identical atomic radii due to lanthanoid contraction.
• Increase in density and hardness: The reduced atomic size leads to higher density and hardness in elements after the lanthanoids.
• Separation difficulty of lanthanoids: The similar radii of lanthanoids make them hard to separate from each other chemically.
• Effect on Basicity of Lanthanide Compounds: The decreasing size of lanthanide ions (from La³⁺ to Lu³⁺) leads to a decrease in their ionic radii, which reduces the basicity of their oxides and hydroxides. For example, La(OH)₃ is more basic than Lu(OH)₃.

Question 4.8: 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 transition elements?

Solution 4.8: Characteristics of Transition Elements:

1. Variable Oxidation States: Transition elements can exhibit multiple oxidation states due to the involvement of d-electrons in bonding.
2. Formation of Colored Compounds: Many transition metal compounds are colored due to d-d electron transitions.
3. Catalytic Properties: They often act as catalysts in various chemical reactions due to their ability to change oxidation states and form complexes.
4. Complex Formation: Transition elements can form complex ions with ligands, leading to a variety of coordination compounds.
5. Magnetic Properties: Some transition metals are paramagnetic due to unpaired d-electrons.
6. High Melting and Boiling Points: They generally have high melting and boiling points compared to other elements.
7. Good Conductors: Transition metals are good conductors of heat and electricity due to the presence of delocalized electrons.

Why Are They Called Transition Elements?

Transition elements are so named because they represent the transition between the s-block and p-block elements in the periodic table. They are found in the d-block of the periodic table and involve the filling of d-orbitals as you move across the series.

d-Block Elements Not Considered Transition Elements:

Not all d-block elements are regarded as transition elements. Specifically, zinc (Zn), cadmium (Cd), and mercury (Hg) are not classified as transition elements.

Reasons: These elements have completely filled d-orbitals in their common oxidation states (Zn²⁺, Cd²⁺, Hg²⁺), making them exhibit very limited variable oxidation states. For example:

• Zinc (Zn): Configuration is [Ar] 3d¹⁰ 4s²; in the Zn²⁺ state, it has a 3d¹⁰ configuration.
• Cadmium (Cd): Configuration is [Kr] 4d¹⁰ 5s²; in the Cd²⁺ state, it has a 4d¹⁰ configuration.
• Mercury (Hg): Configuration is [Xe] 4f¹⁴ 5d¹⁰ 6s²; in the Hg²⁺ state, it has a 5d¹⁰ configuration.

Question 4.9: In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

Solution 4.9: The electronic configuration of transition elements differs from that of non-transition elements in the following ways:

1. D-Orbital Filling:
   • Transition Elements: They have partially filled d-orbitals, with a general configuration of [noble gas] \((n-1)d^{1-10} \, ns^{0-2}\)
   • Non-Transition Elements: They fill their s- and p-orbitals, typically ending in \(ns^{1-2}\) (s-block) or \(ns^{2}np^{1-6} (p-block)\).
2. Variable Oxidation States:
   • Transition Elements: Exhibit multiple oxidation states due to the involvement of d-electrons in bonding.
   • Non-Transition Elements: Generally have fixed oxidation states.
3. Complex Formation:
   • Transition Elements: Readily form coordination complexes due to their ability to accept electron pairs.
   • Non-Transition Elements: Less likely to form such complexes.
4. Colored Compounds:
   • Transition Elements: Often form colored compounds due to d-d electronic transitions.
   • Non-Transition Elements: Typically do not form colored compounds.

Question 4.10: What are the different oxidation states exhibited by lanthanoids?

Solution 4.10: Lanthanides exhibits +2, +3 and +4 oxidation states the most common oxidation state of lanthanoids is +3.

Question 4.11: 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 catalysts.

Solution 4.11: (i) Transition metals and many of their compounds show paramagnetic behaviour because they have unpaired electrons and each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum.

(ii) The enthalpies of atomisation of transition metals are high due to presence of large number of unpaired electrons in their atoms. These atoms have strong interatomic interaction and hence, stronger bonding between them.

(iii) The transition metals generally
form coloured compounds because they partially adsorbed of visible light and Jumps into next orbitals.

(iv) Transition metals are good catalyst because of following reasons:

(a) The ability of transition metal ion to pass easily from ane oxidation state to another and thus providing a new path to reaction with lower activation energy.

(b) The surface of transition metal acts as very good adsorbent and thus provides increased concentration of reactants on their surface causing reaction to occur.

Question 4.12: What are interstitial compounds? Why are such compounds well known for transition metals?

Solution 4.12: Interstitial compounds: The compound in which small atoms like H, C, N, etc. occupies the interstital sites in the crystal lattice are called interstitial compounds.

These compounds are well known for transition metals because small atoms can easily occupy to positions in the voids present in the crystal lattic of transition metals.

Question 4.13: How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Solution 4.13: The variability in oxidation states of transition metals differs significantly from that of non-transition metals due to their electronic structure and bonding characteristics.

Transition Metals:

• Greater Variability: Transition metals have partially filled d-orbitals, allowing them to lose different numbers of electrons.

Examples:

• Iron (Fe): Exhibits oxidation states of +2 (Fe²⁺) and +3 (Fe³⁺).
• Manganese (Mn): Shows multiple oxidation states from +2 to +7, such as +4 (MnO₂) and +7 (KMnO₄).

Non-Transition Metals:

• Limited Variability: Non-transition metals usually have fewer oxidation states, mainly determined by their valence electrons.

Examples:

• Carbon (C): Commonly exhibits +4 (in CO₂) and -4 (in CH₄).
• Sulfur (S): Shows oxidation states like -2 (in H₂S), +4 (in SO₂), and +6 (in SO₄²⁻), but these are fewer than those of transition metals.

In summary, transition metals can exhibit a wide range of oxidation states due to their d-electron involvement, while non-transition metals have more restricted oxidation states primarily based on their s and p-electron configurations.

Question 4.14: Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Solution 4.14: Preparation of Potassium Dichromate from Iron Chromite Ore

The preparation of potassium dichromate \((K_2Cr_2O_7)\) from iron chromite ore \((FeCr_2O_4)\) involves several steps:

1. Roasting: The iron chromite ore is roasted in the presence of an oxidizing agent (usually sodium carbonate \((Na_2CO_3))\) and air. This reaction converts the chromium in the ore to soluble chromate ions \((CrO_4^{2-})\):
   \(4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow\)\( 8Na_2CrO_4 + 4Fe_2O_3 + 8CO_2\)
2. Leaching: The roasted mass is then leached with water to dissolve the sodium chromate formed in the previous step.
3. Precipitation: To convert sodium chromate to potassium dichromate, the sodium chromate solution is treated with potassium chloride (KCl), resulting in the precipitation of potassium dichromate:
   \(Na_2CrO_4 + 2KCl \rightarrow\)\( K_2Cr_2O_7 + 2NaCl\)
4. Crystallization: The resulting solution is concentrated and allowed to crystallize to obtain potassium dichromate crystals.

Effect of Increasing pH on Potassium Dichromate Solution

Increasing the pH of a potassium dichromate solution has significant effects on the chemical equilibrium between chromate \((CrO_4^{2-})\) and dichromate \((Cr_2O_7^{2-})\) ions, described by the following equilibrium:

\(Cr_2O_7^{2-} + 2OH^- \rightleftharpoons\)\( 2CrO_4^{2-} + H_2O\)

• At Low pH: In acidic conditions (low pH), potassium dichromate exists predominantly as dichromate ions \((Cr_2O_7^{2-})\), which gives the solution an orange color.
• At High pH: As the pH increases (becoming more alkaline), the equilibrium shifts to the right, favoring the formation of chromate ions \((CrO_4^{2-})\). This results in a color change from orange (dichromate) to yellow (chromate).

Example:

• Orange Solution: A solution of potassium dichromate at a pH of around 3 appears orange.
• Yellow Solution: Increasing the pH to around 9 or higher causes the solution to turn yellow due to the formation of chromate ions.

This pH-dependent color change is often used in analytical chemistry and demonstrates the reversible nature of the chromate-dichromate system.

Question 4.15: 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

Solution 4.15: Oxidizing Action of Potassium Dichromate

Potassium dichromate \((K_2Cr_2O_7)\) is a strong oxidizing agent, particularly in acidic conditions. The chromium in potassium dichromate is in the +6 oxidation state, and it can be reduced to chromium(III) ions \((Cr^{3+})\), while simultaneously oxidizing other substances. The oxidizing action is often utilized in various redox reactions.

Ionic Equations for Reactions with Potassium Dichromate

(i) Reaction with Iodide

When potassium dichromate reacts with iodide ions \((I^-)\), it oxidizes iodide to iodine \((I_2)\), and itself gets reduced to chromium(III) ions.

Ionic Equation:
\(Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow\)\( 2Cr^{3+} + 3I_2 + 7H_2O\)

(ii) Reaction with Iron(II) Solution

In the reaction with iron(II) ions \((Fe^{2+})\), potassium dichromate oxidizes Fe(II) to Fe(III), while being reduced to Cr(III).

Ionic Equation:
\(Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow\)\( 2Cr^{3+} + 6Fe^{3+} + 7H_2O\)

(iii) Reaction with Hydrogen Sulfide

When potassium dichromate reacts with hydrogen sulfide \((H_2S)\), it oxidizes H₂S to sulfur (S) or sulfate ions \((SO_4^{2-})\), depending on the reaction conditions, while being reduced to chromium(III) ions.

Ionic Equation (for oxidation to elemental sulfur):
\(Cr_2O_7^{2-} + 3H_2S + 14H^+ \rightarrow\)\( 2Cr^{3+} + 3S + 7H_2O\)

Question 4.16: 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.

Solution 4.16: Preparation of Potassium Permanganate

Potassium permanganate \((KMnO_4)\) is primarily prepared through the oxidation of manganese(II) compounds, usually manganese dioxide \((MnO_2)\), using potassium hydroxide \((KOH)\) and potassium chlorate \((KClO_3)\) or by other chemical processes. Here is a common method:

1. Manganese Dioxide with Potassium Hydroxide:

• Manganese dioxide is reacted with potassium hydroxide and oxidizing agents such as chlorine gas or potassium chlorate. The general reaction can be represented as:
  \(3MnO_2 + 2KOH + 2ClO_3^- \rightarrow\)\( 2KMnO_4 + 3H_2O + 2Cl^-\)

2. Crystallization: The resulting solution is filtered and concentrated, and potassium permanganate crystals are obtained upon cooling.

Reactions of Acidified Potassium Permanganate Solution

Acidified potassium permanganate solution is a strong oxidizing agent, and it reacts with several substances. The acidic medium is usually provided by sulfuric acid \((H_2SO_4)\).

(i) Reaction with Iron(II) Ions

When permanganate ions react with iron(II) ions \((Fe^{2+})\) in an acidic solution, the iron(II) ions are oxidized to iron(III) ions \((Fe^{3+})\), while the permanganate ions are reduced to manganese(II) ions \((Mn^{2+})\).

Ionic Equation:
\(MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow\)\( Mn^{2+} + 4H_2O + 5Fe^{3+}\)

(ii) Reaction with Sulfur Dioxide

In an acidic medium, sulfur dioxide \((SO_2)\) can react with permanganate ions, oxidizing SO₂ to sulfuric acid \((H_2SO_4)\), while the permanganate ions are reduced to manganese(II) ions.

Ionic Equation:
\(2MnO_4^- + 5SO_2 + 6H^+ \rightarrow\)\( 2Mn^{2+} + 5H_2SO_4\)

(iii) Reaction with Oxalic Acid

When oxalic acid \((H_2C_2O_4)\) reacts with acidified potassium permanganate, the oxalic acid is oxidized to carbon dioxide \((CO_2)\), and the permanganate ions are reduced to manganese(II) ions.

Ionic Equation:
\(2MnO_4^- + 16H^+ + 5H_2C_2O_4 \rightarrow\)\( 2Mn^{2+} + 10CO_2 + 8H_2O\)

Question 4.17: For M²⁺/M and M³⁺/M²⁺ systems, the E° values for some metals are as follows:
Cr²⁺/Cr -0.9V  Cr³/Cr²⁺ -0.4 V
Mn²⁺/Mn -1.2V  Mn³⁺/Mn²⁺ +1.5 V
Fe²⁺/Fe -0.4V Fe³⁺/Fe²⁺ +0.8 V 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.

Solution 4.17:
(i) The reduction potentials for the given pairs increase in the following order.

Mn²⁺ / Mn < Cr²⁺ / Cr < Fe²⁺ /Fe

So, the oxidation of Fe to Fe²⁺ is not as easy as the oxidation of Cr to Cr²⁺ and the oxidation of Mn to Mn²⁺. Thus, these metals can be arranged in the increasing order of their ability to get oxidised as Fe < Cr < Mn

(ii) The E° value for Fe³⁺/Fe²⁺ is higher than that for Cr³⁺/Cr²⁺ and lower than that for Mn³⁺/Mn²⁺. So, the reduction of Fe³⁺ to Fe²⁺ is easier than the reduction of Mn³⁺ to Mn²⁺, but not as easy as the reduction of Cr³⁺ to Cr²⁺. Hence, Fe³⁺ is more stable than Mn³⁺, but less stable than Cr³⁺. These metal ions can be arranged in the increasing order of their stability as Mn³⁺ < Fe³⁺ < Cr³⁺.

Question 4.18: Predict which of the following will be coloured in an aqueous solution. Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.

Solution 4.18: Ions with electrons in the d-orbital will be the only ones that will be coloured, and the ions with empty d-orbital will be colourless.

Element	Atomic Number	Ionic State	Electronic configuration in ionic state
Ti	22	T13+	[Ar] 3d1
V	23	V3+	[Ar] 3d2
Cu	29	Cu+	[Ar] 3d10
Sc	21	Sc3+	[Ar]
Mn	25	Mn2+	[Ar] 3d5
Fe	26	Fe3+	[Ar] 3d5
Co	27	Co2+	[Ar] 3d7

All other ions, except Sc³⁺, will be coloured in an aqueous solution because of d−d transitions. Not Sc^3+, as it has an empty d-orbital.

Question 4.19: Compare the stability of the +2 oxidation state for the elements of the first transition series.

Solution 4.19:

Sc			+3
Ti	+1	+2	+3	+4
V	+1	+2	+3	+4	+5
Cr	+1	+2	+3	+4	+5	+6
Mn	+1	+2	+3	+4	+5	+6	+7
Fe	+1	+2	+3	+4	+5	+6
Co	+1	+2	+3	+4	+5
Ni	+1	+2	+3	+4
Cu	+1	+2	+3
Zn		+2

From the above table, it is evident that the maximum number of oxidation states is shown by Mn, varying from +2 to +7. The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.

Question 4.20: 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

Solution 4.20: (i) Electronic configuration: The general electronic configuration for lanthanoids is [Xe]⁵⁴ 4f^0-14 5d^0-1 6s^2, and that for actinoids is [Rn]⁸⁶ 5f^1-14 6d^0-1 7s². Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Atomic and ionic sizes: Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.

(iii) Oxidation states: The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of the extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in the +3 state than in the +4 state.

(iv) Chemical reactivity: In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similarly to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most non-metals at moderate temperatures. Alkalies have no action on these actinoids. In the case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Question 4.21: How would you account for the following?
(i) Of the d4 species, Cr²⁺ is strongly reducing while manganese(III) is strongly oxidising.
(ii) Cobalt(II) is stable in an aqueous solution, but in the presence of complexing reagents, it is easily oxidised.
(iii) The d¹ configuration is very unstable in ions.

Solution 4.21:

(i) Cr²⁺ is strongly reducing in nature. It has a d⁴ configuration. While acting as a reducing agent, it gets oxidised to Cr³⁺ (electronic configuration, d³). 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) Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidised to Co (III). Although the 3rd ionisation energy for Co is high, the higher amount of crystal field stabilisation energy (CFSE) released in the presence of strong-field ligands overcomes this ionisation energy.

(iii) The ions in the d¹ configuration tend to lose one more electron to get into a stable d⁰ configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

Question 4.22: What is meant by ‘disproportionation’? Give two examples of disproportionation reactions in an aqueous solution.

Solution 4.22: Disproportionation is a specific type of redox reaction in which a single substance is simultaneously oxidized and reduced, resulting in the formation of two different products. In this process, one species undergoes an increase in oxidation state (oxidation) while another portion of the same species undergoes a decrease in oxidation state (reduction).

Examples of Disproportionation Reactions in Aqueous Solution

1. Disproportionation of Hydrogen Peroxide

When hydrogen peroxide \((H_2O_2)\) is mixed with an acidic solution, it can disproportionate into water and oxygen gas:

\(2H_2O_2 \rightarrow 2H_2O + O_2\)

• Oxidation: One molecule of hydrogen peroxide is oxidized (loses electrons) to form oxygen (O₂).
• Reduction: Another molecule of hydrogen peroxide is reduced (gains electrons) to form water (H₂O).

2. Disproportionation of Chlorine in Water

Chlorine gas (Cl₂) when dissolved in water, can undergo disproportionation to form hydrochloric acid (HCl) and hypochlorous acid (HClO):

\(Cl_2 + H_2O \rightarrow HCl + HClO\)

• Oxidation: One chlorine atom is oxidized to hypochlorous acid (HClO).
• Reduction: The other chlorine atom is reduced to hydrochloric acid (HCl).

Question 4.23: Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Solution 4.23: In the first series of transition metals, Cu exhibits +1 oxidation state very frequently. We know Cu has an electronic configuration of [Ar] 3d ¹⁰, that is, the completely filled d-orbital makes it highly stable.

Question 4.24: 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 an aqueous solution?

Solution 4.24:

	Gaseous ions	Number of unpaired electrons
1	Mn3+,[Ar]3d4	4
2	Cr3+,[Ar]3d3	3
3	V3+,[Ar]3d2	2
4	Ti3+,[Ar]3d1	1

Cr³⁺ is the most stable in aqueous solutions owing to a t_2g³ configuration.

Question 4.25: Give examples and suggest reasons for the following features of transition metal chemistry.
(i) The lowest oxide of transition metal is basic, and the highest is amphoteric/acidic.
(ii) A transition metal exhibits the highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Solution 4.25: (i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base. On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding, so they are unavailable. There is also a highly effective nuclear charge. As a result, it can accept electrons and behave as an acid.

For example, Mn^IIO is basic and Mn₂^VIIO₇is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides.

For example, in OsF₆ and V₂O₅, the oxidation states of Os and V are +6 and +5, respectively.

(iii) 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, the oxidation state of Mn is +7.

Question 4.26: Indicate the steps in the preparation of: (i) K₂Cr₂O₇ from chromite ore. (ii) KMnO₄ from pyrolusite ore.

Solution 4.26: Preparation of K₂Cr₂O₇ from Chromite Ore

The preparation of potassium dichromate \((K_2Cr_2O_7)\) from chromite ore \((FeCr_2O_4)\) involves the following steps:

1. Roasting: The chromite ore is roasted in the presence of sodium carbonate \((Na_2CO_3)\) and an oxidizing agent (such as air or oxygen). This process converts chromium(II) in the ore to soluble chromate ions.

• Reaction:
  \(4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow\)\( 8Na_2CrO_4 + 4Fe_2O_3 + 8CO_2\)

2. Leaching: The roasted mass is leached with water to dissolve the sodium chromate \((Na_2CrO_4)\) formed in the previous step.

3. Precipitation: The sodium chromate solution is treated with potassium chloride (KCl) to precipitate potassium dichromate.

• Reaction:
  \(Na_2CrO_4 + 2KCl \rightarrow\)\( K_2Cr_2O_7 + 2NaCl\)

4. Crystallization: The resulting potassium dichromate solution is concentrated and allowed to crystallize to obtain pure potassium dichromate crystals.

Preparation of KMnO₄ from Pyrolusite Ore

The preparation of potassium permanganate (KMnO₄) from pyrolusite ore (MnO₂) involves the following steps:

1. Oxidation of Manganese Dioxide: Pyrolusite is reacted with potassium hydroxide (KOH) and oxidizing agents (usually oxygen or potassium permanganate itself) in an alkaline medium. This process forms potassium permanganate.

• Reaction:
  \(3MnO_2 + 4KOH + 2H_2O \rightarrow\)\( 2KMnO_4 + 3H_2O_2\)

2. Neutralization: The resulting solution containing potassium permanganate is neutralized, often using an acid, to isolate the compound.

3. Crystallization: The concentrated solution is evaporated, allowing the formation of potassium permanganate crystals.

Question 4.27: What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Solution 4.27: Alloys are materials composed of two or more elements, with at least one of them being a metal. The resulting mixture typically exhibits enhanced properties, such as increased strength, improved corrosion resistance, or better electrical conductivity, compared to the individual components.

Important Alloy Containing Lanthanoid Metals

An important alloy that contains lanthanoid metals is neodymium-iron-boron (NdFeB).

Composition:

• Neodymium (Nd): A lanthanoid metal that contributes to the alloy’s magnetic properties.
• Iron (Fe): Provides strength and stability.
• Boron (B): Enhances magnetic properties.

Uses of Neodymium-Iron-Boron Alloy: Permanent Magnets: NdFeB alloys are used to create high-performance permanent magnets. These magnets are found in various applications, including:

• Electric Motors: Used in electric vehicles, industrial machinery, and household appliances.
• Magnetic Resonance Imaging (MRI): Utilized in MRI machines for medical imaging.
• Speakers and Headphones: Provide powerful sound in audio equipment.
• Computer Hard Drives: Enhance data storage capabilities.

The combination of neodymium, iron, and boron results in one of the strongest permanent magnets available, making it a vital material in modern technology.

Question 4.28: 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.

Solution 4.28: Inner transition metals: Elements in which the last electron enters the f-orbital.

The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements.

From the given atomic numbers, the inner transition elements are the ones with atomic numbers 59, 95, and 102.

Question 4.29: The chemistry of the actinoid elements is not as smooth as that of the lanthanoids. Justify this statement by giving some examples of the oxidation state of these elements.

Solution 4.29: Lanthanoids project 3 oxidation states (+2, +3, +4). From these, the +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. While the energy difference between 5f, 6d, and 7s orbitals’ is very less. Hence, actinoids display a large number of oxidation states.

For example, neptunium displays +3, +4, +5, and +7, while uranium and plutonium display +3, +4, +5, and +6 oxidation states. The most common oxidation state in the case of actinoids is also +3.

Question 4.30: Which is the last element in the series of actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Solution 4.30: In the actinoid series, the last element is lawrencium, Lr. The atomic number of the element is 103, and its electronic configuration is [Rn] 5f¹⁴ 6d¹ 7s². The most common oxidation state displayed by it is +3; because after losing 3 electrons, it attains a stable f¹⁴ configuration.

Question 4.31: Use Hund’s rule to derive the electronic configuration of the Ce³⁺ ion, and calculate its magnetic moment on the basis of the ‘spin-only’ formula.

Solution 4.31:

Ce: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹5d¹6s²

Ce³⁺: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹

Magnetic moment can be calculated as \(\nu = \sqrt{n(n+2)}\)

Where,

n = number of unpaired electrons

In Ce³⁺, n = 1

\(\nu = \sqrt{1(1+2)}\)

\({l}= \sqrt{3)}\)

= 1. 732 BM

Question 4.32: 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.

Solution 4.32: The lanthanides that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.

+2	+4
Nd (60)	Ce (58)
Sm (62)	Pr (59)
Eu (63)	Nd (60)
Tm (69)	Tb (65)
Yb (70)	Dy (66)

In the parenthesis are the atomic numbers of the elements given.

Tb, after forming Tb⁴⁺, attains a stable electronic configuration of [Xe] 4f⁷.

Yb, after forming Yb²⁺, attains a stable electronic configuration of [Xe] 4f¹⁴.

Eu, after forming Eu²⁺, attains a stable electronic configuration of [Xe] 4f⁷.

Ce, after forming Ce⁴⁺, attains a stable electronic configuration of [Xe].

Question 4.33: Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) chemical reactivity. (ii) electronic configuration (iii) oxidation states

Solution 4.33: The actinoids and lanthanoids are two series of f-block elements in the periodic table. They exhibit distinct differences in their chemistry, which can be summarized in the following points:

(i) Chemical Reactivity

• Lanthanoids:
  • The lanthanoids generally have moderate reactivity, increasing from lanthanum to lutetium.
  • They react with water and acids, but are less reactive compared to the actinoids. For example, they react with dilute acids to release hydrogen gas.
  • Their reactivity can be attributed to the presence of f-electrons, which can influence bonding.
• Actinoids:
  • The actinoids are typically more reactive than lanthanoids.
  • They can react vigorously with water, acids, and even oxygen, often forming oxides and hydrides.
  • Their higher reactivity is due to their larger atomic size and the greater energy of their f-orbitals, which facilitates reactions with a wider range of substances.

(ii) Electronic Configuration

• Lanthanoids:
  • The lanthanoids have the general electronic configuration of:
    \([Xe] 6s^2 4f^n \quad (n = 1 \text{ to } 14)\)
  • In this series, the 4f subshell is progressively filled as one moves from cerium (Ce) to lutetium (Lu).
• Actinoids:
  • The actinoids have the general electronic configuration of:
    \([Rn] 7s^2 5f^n \quad (n = 1 \text{ to } 14)\)
  • In this series, the 5f subshell is progressively filled as one moves from thorium (Th) to lawrencium (Lr).

(iii) Oxidation States

• Lanthanoids:
  • The most common oxidation state of lanthanoids is +3.
  • However, some elements can exhibit +2 or +4 oxidation states, but these are less common (e.g., cerium in +4 and samarium in +2).
  • The stability of the +3 oxidation state is due to the half-filled f-orbitals.
• Actinoids:
  • Actinoids exhibit a wider range of oxidation states, typically ranging from +3 to +6.
  • The most common oxidation state is +3, but many actinoids can exhibit +4 (e.g., uranium, plutonium) and +5 (e.g., neptunium, americium) due to their ability to lose f-electrons.
  • Uranium, for instance, can exist in the +6 oxidation state as uranyl \((UO_2^{2+})\).

Summary of Comparison

Aspect	Lanthanoids	Actinoids
Chemical Reactivity	Moderate, increases across the series	High, generally more reactive than lanthanoids
Electronic Configuration	\([Xe] 6s^2 4f^n\)	\([Rn] 7s^2 5f^n\)
Oxidation States	Commonly +3, some +2 and +4	Range from +3 to +6, commonly +3, +4, +5, and +6

Question 4.34: Write the electronic configurations of the elements with the atomic numbers 61, 91, 101 and 109.

Solution 4.34: Here are the electronic configurations for the elements with atomic numbers 61, 91, 101, and 109:

1. Element with Atomic Number 61: Promethium (Pm)

• Electronic Configuration: \([Xe] 6s^2 4f^5\)

2. Element with Atomic Number 91: Protactinium (Pa)

• Electronic Configuration: \([Rn] 7s^2 5f^2\)

3. Element with Atomic Number 101: Mendelevium (Md)

• Electronic Configuration: \([Rn] 7s^2 5f^{13}\)

4. Element with Atomic Number 109: Meitnerium (Mt)

• Electronic Configuration: \([Rn] 7s^2 5f^{14} 6d^7\)

• Promethium (Pm, 61): \([Xe] 6s^2 4f^5\)
• Protactinium (Pa, 91): \([Rn] 7s^2 5f^2\)
• Mendelevium (Md, 101): \([Rn] 7s^2 5f^{13}\)
• Meitnerium (Mt, 109): \([Rn] 7s^2 5f^{14} 6d^7\)

Question 4.35: Compare the general characteristics of the first series of 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

Solution 4.35: (i) In the 1st, 2nd and 3rd transition series, the 3d, 4d, and 5d orbitals are respectively filled. We know that elements in the same vertical column generally have similar electronic configurations.

In the first transition series, two elements show unusual electronic configurations

Cu (29) = 3d¹⁰4s¹

Cr (24) = 3d⁵4s¹

Similarly, there are exceptions in the second transition series. These are

Ag(47) = 4d¹⁰5s¹

Pd(46) = 4d¹⁰ 5s⁰

Rh(45) = 4d⁸5s¹

Ru (44) = 4d⁷5s¹

Tc (43) = 4d⁶5s¹

Mo (42) = 4d⁵5s¹

There are some exceptions in the third transition series as well. These are

Au (79 ) = 5d¹⁰ 6s¹

Pt (78) = 5d⁹ 6s¹

W (74) = 5d⁴ 6s²

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

(ii) In each of the three transition series, the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends. However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and third transition series, wherein higher oxidation states are more important.

Some stable complexes are

\({\left [ Fe^{II}(Cn)_6 \right ]}^{4-}\), \({\left [ Co^{III}(NH_3)_6 \right ]}^{3+}\), \({\left [ Ti(H_2O)_6 \right ]}^{3+}\)

The issue is that no such complexes are known for the second and third transition series, such as Mo, W, Rh, In. They form complexes in which their oxidation states are high.

For example: WCl₆, ReF₇, RuO₄, etc.

(iii) In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series. Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the 2nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1^st transition series.

(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, the atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series. However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

Question 4.36: Write down the number of 3d electrons in each of the following ions. Ti²⁺, 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).

Solution 4.36:

Metal ion	Number of d-electrons	Filling of d-orbitals
Ti2+	2	t2g2
V2+	3	t2g3
Cr3+	3	t2g3
Mn2+	5	t2g3 eg2
Fe2+	6	t2g4 eg2
Fe3+	5	t2g3 eg2
CO2+	7	t2g5 eg2
Ni2+	8	t2g6 eg2
Cu2+	9	t2g6 eg3

Question 4.37: Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Solution 4.37: In many ways, the properties of elements of heavier transition elements differ from those of the first transition series.

(a) The elements of the first transition series form low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

(b) The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.

(c) 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 stronger metallic bonding (M−M bonding).

(d)  For heavier elements, the higher oxidation states are more common, whereas, for first transition series elements, th +2 and +3 oxidation states are more common.

(e) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series). However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

Question 4.38: What can be inferred from the magnetic moment values of the following complex species?

Example	Magnetic Moment (BM)
K4[Mn(CN)6]	2.2
[Fe(H2O)6]2+	5.3
K2[MnCl4]	5.9

Solution 4.38: Magnetic moment (μ) is given as:

\(\nu = \sqrt{n(n+2)}\)

For value n = 1,

\(\nu = \sqrt{1(1+2)} = \sqrt{3} = 1.732\)

For value n = 2,

\(\nu = \sqrt{2(2+2)} = \sqrt{8} = 2.83\)

For value n = 3,

\(\nu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\)

For value n = 4,

\(\nu = \sqrt{4(4+2)} = \sqrt{24} = 4.899\)

For value n = 5,\

\(\nu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\)

K₄[Mn(CN)₆]

\(\sqrt{n(n+2)} = 2.2\)

For in transition metals, the magnetic moment is calculated from the spin-only formula.

Therefore, we can see from the above calculation that the given value is closest to n = 1. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that CN⁻ is a strong field ligand that causes the pairing of electrons.

[Fe(H₂O)₆]²⁺

\(\sqrt{n(n+2)} = 5.3\)

Thus, the calculation given above proves that the closest value to n= 4. In addition to this, the +2 oxidation state in this complex is none other than Fe. The d-orbital for Fe would have 6 electrons.

Thus, we can assume that H2O is a weak field ligand and does not induce electron pairing.

K₂[MnCl₄] 

\(\sqrt{n(n+2)} = 5.9\)

From the equation, we can see that the given value is nearest to n = 5. Mn is in the +2 oxidation state in this complex. This implies that in the d-orbital, Mn has 5 electrons.

We may also assume that Cl – is a weak field ligand and does not induce electron pairing.