NCERT Solutions for Class 12: D and F Block Elements Chapter 4

For students aiming for excellence in their Class 12 exams, mastering “D and F Block Elements” is crucial. Our NCERT Solutions for Chapter 4, titled “D and F Block Elements Class 12,” are designed to provide a thorough understanding of this essential topic. These solutions align with the latest syllabus for 2024-25, ensuring you get accurate and up-to-date answers.

The chapter on D and F Block Elements is pivotal for both board exams and various entrance tests, making it indispensable for your academic success. By practicing with our detailed solutions, you can reinforce your knowledge and increase your confidence in tackling exam questions.

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Question 1. Write down the electronic configuration of:

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

Solution:

(i) Cr³⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
Or, [Ar] 3d³

(ii) Pm³⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f³
Or, [Xe] 4f³

(iii) Cu⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
Or, [Ar] 3d¹⁰

(iv) Ce⁴⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶
Or, [Xe]

(v) Co²⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷
Or, [Ar] 3d⁷

(vi) Lu²⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹
Or, [Xe] 4f¹⁴ 5d¹

(vii) Mn²⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Or, [Ar] 3d⁵

(viii) Th⁴⁺
Electronic Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s²
Or, [Rn] 6d²

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

Solution:

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

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

It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in the +2 state has a stable 3d⁵ configuration. This is the reason Mn²⁺ shows resistance to oxidation to Mn³⁺. Also, Fe²⁺ has a 3d⁶ configuration and by losing one electron, its configuration changes to a more stable 3d⁵ configuration. Therefore, Fe²⁺ easily gets oxidized to Fe³⁺ oxidation state.

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

Solution:

The oxidation states displayed by the first half of the first row of transition metals are given in the table below:

	Sc	Ti	V	Cr	Mn
+2	+2	+2	+2	+2	+2
+3	+3	+3	+3	+3	+3
+4	+4	+4	+4
+5			+5		+5
+6					+6
+7

It can be easily observed that, except for Sc, all other metals display the +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.

• Sc (+2) = 3d¹
• Ti (+2) = 3d²
• V (+2) = 3d³
• Cr (+2) = 3d⁴
• Mn (+2) = 3d⁵

The +2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in the +2 state also increases from Ti(+2) to Mn(+2), the stability of the +2 state increases (as the d-orbital becomes more and more half-filled). Mn(+2) has a d⁵ configuration (a half-filled d shell), which is highly stable.

Question 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:

The elements in the first half of the transition series exhibit many oxidation states, with Mn showing the maximum number of oxidation states ranging from +2 to +7. The stability of the +2 oxidation state increases with the increase in atomic number. This occurs because more electrons are getting filled in the d-orbital.

• Scandium (Sc) does not show the +2 oxidation state. Its electronic configuration is 4s² 3d¹. It loses all three electrons to form Sc³⁺. The +3 oxidation state of Sc is very stable because, by losing all three electrons, it attains a stable noble gas configuration, [Ar].
• Titanium (Ti) and Vanadium (V) show stable +4 and +5 oxidation states, respectively, for similar reasons. After losing electrons, these elements achieve stable configurations that are favored energetically.
• Manganese (Mn) has a very stable +2 oxidation state. After losing two electrons, its d-orbital is exactly half-filled, resulting in a stable [Ar] 3d⁵ configuration.

The stability of these oxidation states is significantly influenced by the electronic configurations, particularly the stability associated with half-filled and fully filled d-orbitals.

Question 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:

Electronic Configuration in Ground State	Stable Oxidation States
(i) 3d³ (Vanadium)	+2, +3, +4, +5
(ii) 3d⁵ (Chromium)	+3, +4, +6
(iii) 3d⁵ (Manganese)	+2, +4, +6, +7
(iv) 3d⁸ (Cobalt)	+2, +3
(v) 3d⁴	There is no 3d⁴ configuration in the ground state.

Question 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.

Solution:

(i) Vanadate
Oxidation state of V is +5.

(ii) Chromate
Oxidation state of Cr is +6.

(iii) Permanganate
Oxidation state of Mn is +7.

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

Solution:

Lanthanoid Contraction:

As we move along the lanthanoid series, the atomic number increases gradually by one for each element. This means that the number of electrons and protons in the atom also increases by one. Since electrons are added to the same 4f shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of a proton is more pronounced than the increase in the interelectronic repulsions due to the addition of an electron. Additionally, the 4f electrons have a poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases, leading to a steady decrease in the size of lanthanoids with increasing atomic number. This phenomenon is known as lanthanoid contraction.

Consequences of Lanthanoid Contraction:

(i) There is a similarity in the properties of the second and third transition series due to lanthanoid contraction.

(ii) The separation of lanthanoids is made possible because of lanthanoid contraction.

(iii) Lanthanoid contraction also causes variation in the basic strength of lanthanide hydroxides. The basic strength decreases from La(OH)₃ to Lu(OH)₃.

Question 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?

Solution:

Characteristics of Transition Elements:

1. Partially Filled d-Orbital: Transition elements are those elements in which the atoms or ions (in stable oxidation states) contain partially filled d-orbitals.
2. Variable Oxidation States: They exhibit a variety of oxidation states, typically differing by 1 or 2 units.

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

Solution:

In the lanthanide series, the most common oxidation state is +3, i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in solutions or solid compounds.

Question 11. Explain giving reasons:

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

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

(iii) The transition metals generally form colored compounds.

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

Solution:

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

Transition metals show paramagnetic behavior due to the presence of unpaired electrons in their d-orbitals. Each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum. Even though the orbital angular momentum is quenched in the first transition series, the unpaired electrons contribute to the overall paramagnetism of the metal.

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

Transition metals have high effective nuclear charge and a large number of valence electrons, which result in strong metallic bonds. The strong bonding between metal atoms requires a significant amount of energy to break these bonds, leading to high enthalpies of atomization for transition metals.

(iii) The transition metals generally form colored compounds.

Transition metals form colored compounds due to the absorption of visible light, which promotes electrons from one d-orbital to another. In the presence of ligands, the d-orbitals split into two sets with different energies. The energy required for these d-d transitions falls within the visible light spectrum. The ions of transition metals absorb specific wavelengths of light, and the remaining light is reflected, giving the compound its color.

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

Transition metals are effective catalysts due to two main reasons:

(a) Variable Oxidation States: They can exhibit different oxidation states and form complexes, allowing them to create unstable intermediate compounds. This ability provides an alternative reaction pathway with a lower activation energy.

(b) Suitable Surface: Transition metals offer a suitable surface for reactions to occur, facilitating the adsorption of reactants and the subsequent release of products. This enhances their catalytic activity.

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

Solution:

Interstitial compounds are compounds in which small atoms, such as hydrogen, carbon, or nitrogen, occupy the interstitial sites (small voids) between the larger metal atoms in the crystal lattice of a metal. Transition metals are well known for forming such compounds because they have large atomic sizes and numerous interstitial sites in their crystal structures. This allows them to effectively trap and bond with small atoms, leading to the formation of interstitial compounds.

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

Solution:

In transition metals, the oxidation states can vary widely, ranging from +1 up to the highest oxidation state by removing all its valence electrons. Additionally, the oxidation states of transition metals can differ by just one unit. For example, iron can exist in the +2 (Fe²⁺) and +3 (Fe³⁺) oxidation states, and copper can exist in the +1 (Cu⁺) and +2 (Cu²⁺) states.

In contrast, non-transition metals typically show a smaller range of oxidation states, often differing by two units. For example, in non-transition elements, the oxidation states can differ by 2, such as +2 and +4 or +3 and +5.

Question 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:

Preparation of Potassium Dichromate from Iron Chromite Ore:

1. Preparation of Sodium Chromate:
   • Iron chromite ore (FeCr₂O₄) is roasted with sodium carbonate (Na₂CO₃) in the presence of air to form sodium chromate (Na₂CrO₄).
2. Conversion of Sodium Chromate into Sodium Dichromate:
   • Sodium chromate is then treated with hydrochloric acid (HCl) to convert it into sodium dichromate (Na₂Cr₂O₇).
3. Conversion of Sodium Dichromate to Potassium Dichromate:
   • Sodium dichromate is then treated with potassium chloride (KCl) to produce potassium dichromate (K₂Cr₂O₇). Potassium dichromate is less soluble in water compared to sodium chloride, so it crystallizes out and can be collected by filtration.

Effect of Increasing pH on a Solution of Potassium Dichromate:

Potassium dichromate (K₂Cr₂O₇) in solution exists in equilibrium with chromate (CrO₄²⁻) and dichromate (Cr₂O₇²⁻) ions depending on the pH of the solution. The equilibrium can be represented as:

• At low pH (acidic conditions):
  $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightleftharpoons 2\text{HCrO}_4^- + 6\text{H}_2\text{O}$ 

  Cr2​O72−​+14H+⇌2HCrO4−​+6H2​O

• At high pH (basic conditions):
  $2\text{CrO}_4^{2-} + 2\text{H}_2\text{O} \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^-$ 

  2CrO42−​+2H2​O⇌Cr2​O72−​+2OH−

As the pH of the solution increases, the equilibrium shifts towards the formation of chromate ions (CrO₄²⁻) from dichromate ions (Cr₂O₇²⁻).

Question 15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i) iodide

(ii) iron(II) solution

(iii) H₂S

Solution:

Potassium dichromate (K₂Cr₂O₇) acts as a very strong oxidizing agent, especially in acidic medium. It accepts electrons and gets reduced in the process, thereby oxidizing other substances.

(i) Reaction with iodide:

Potassium dichromate oxidizes iodide ions (I⁻) to iodine (I₂). The ionic equation for this reaction is:

 

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}$

Cr2​O72−​+14H++6I−→2Cr3++3I2​+7H2​O

(ii) Reaction with iron(II) solution:

Potassium dichromate oxidizes iron(II) ions (Fe²⁺) to iron(III) ions (Fe³⁺). The ionic equation for this reaction is:

 

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$

Cr2​O72−​+14H++6Fe2+→2Cr3++6Fe3++7H2​O

(iii) Reaction with H₂S:

Potassium dichromate oxidizes hydrogen sulfide (H₂S) to sulfur (S). The ionic equation for this reaction is:

 

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 3\text{S} + 7\text{H}_2\text{O}$

Cr2​O72−​+14H++3H2​S→2Cr3++3S+7H2​O

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

(i) iron(II) ions

(ii) SO₂

(iii) oxalic acid

Solution:

Preparation of Potassium Permanganate:

Potassium permanganate (KMnO₄) can be prepared from pyrolusite (MnO₂) using the following steps:

1. Fusing the Ore: Pyrolusite (MnO₂) is fused with potassium hydroxide (KOH) in the presence of atmospheric oxygen or an oxidizing agent like potassium nitrate (KNO₃) or potassium perchlorate (KClO₄). This reaction yields potassium manganate (K₂MnO₄), which is green in color.
2. Extraction and Oxidation:
   • The green mass of potassium manganate (K₂MnO₄) is extracted with water.
   • This manganate solution is then oxidized to potassium permanganate (KMnO₄) either electrolytically or by passing chlorine (Cl₂) or ozone (O₃) through it.
   • Electrolytic Oxidation: At the anode, manganate ions (MnO₄²⁻) are oxidized to permanganate ions (MnO₄⁻).
   • Oxidation by Chlorine: Chlorine gas (Cl₂) is passed through the manganate solution to convert it to permanganate.
   • Oxidation by Ozone: Ozone gas (O₃) is used for the same purpose.

Reactions of Acidified Permanganate Solution:

(i) Reaction with Iron(II) Ions:

Acidified potassium permanganate (KMnO₄) oxidizes iron(II) ions (Fe²⁺) to iron(III) ions (Fe³⁺). The ionic equation is:

 

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

MnO4−​+8H++5Fe2+→Mn2++5Fe3++4H2​O

(ii) Reaction with SO₂:

Acidified potassium permanganate oxidizes sulfur dioxide (SO₂) to sulfuric acid (H₂SO₄). The ionic equation is:

 

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{SO}_2 \rightarrow \text{Mn}^{2+} + 5\text{SO}_4^{2-} + 4\text{H}_2\text{O}$

MnO4−​+8H++5SO2​→Mn2++5SO42−​+4H2​O

(iii) Reaction with Oxalic Acid:

Acidified potassium permanganate oxidizes oxalic acid (H₂C₂O₄) to carbon dioxide (CO₂). The ionic equation is:

 

$\text{MnO}_4^- + 16\text{H}^+ + 5\text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$

MnO4−​+16H++5H2​C2​O4​→Mn2++10CO2​+8H2​O

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

Cr²⁺/Cr = -0.9 V
Cr³⁺/Cr²⁺ = -0.4 V

Mn²⁺/Mn = -1.2 V
Mn³⁺/Mn²⁺ = +1.5 V

Fe²⁺/Fe = -0.4 V
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⁺

(ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Solution:

(i) Stability of Fe³⁺:

The reduction potentials for the given pairs indicate:

• Fe³⁺/Fe²⁺ = +0.8 V
• Cr³⁺/Cr²⁺ = -0.4 V
• Mn³⁺/Mn²⁺ = +1.5 V

The value for Fe³⁺/Fe²⁺ is higher than Cr³⁺/Cr²⁺ and lower than Mn³⁺/Mn²⁺. Thus, 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⁺. The metal ions can be arranged in the increasing order of their stability as: Mn³⁺ < Fe³⁺ < Cr³⁺

(ii) Ease of Oxidation:

The reduction potentials for the given pairs indicate:

• Mn²⁺/Mn = -1.2 V
• Cr²⁺/Cr = -0.9 V
• Fe²⁺/Fe = -0.4 V

The lower the reduction potential, the easier it is for the metal to be oxidized. Therefore, the oxidation of Fe to Fe²⁺ is less easy compared to the oxidation of Cr to Cr²⁺ and Mn to Mn²⁺. The metals can be arranged in the increasing order of their ability to get oxidized as: Fe < Cr < Mn

Question 18. Predict which of the following ions will be colored in aqueous solution:

$\text{Ti}^{3+}$

Ti3+,

$\text{V}^{3+}$

V3+,

$\text{Cu}^{+}$

Cu+,

$\text{Sc}^{3+}$

Sc3+,

$\text{Mn}^{2+}$

Mn2+,

$\text{Fe}^{3+}$

Fe3+, and

$\text{Co}^{2+}$

Co2+. Give reasons for each.

Solution:

Transition metal ions exhibit color in aqueous solutions due to electronic transitions, specifically d-d transitions that occur within their incompletely filled d orbitals. The color arises when electrons are promoted from lower energy d-orbitals to higher energy ones within the same subshell.

Colored Ions:

1. 
   $\text{Ti}^{3+}$ 

   Ti3+: Has the electronic configuration [Ar] 3d¹. The single electron in the d-orbital can undergo d-d transitions, making it colored.

2. 
   $\text{V}^{3+}$ 

   V3+: Has the electronic configuration [Ar] 3d². The two electrons in the d-orbitals can undergo d-d transitions, making it colored.

3. 
   $\text{Mn}^{2+}$ 

   Mn2+: Has the electronic configuration [Ar] 3d⁵. With five unpaired electrons, it can exhibit d-d transitions, making it colored.

4. 
   $\text{Fe}^{3+}$ 

   Fe3+: Has the electronic configuration [Ar] 3d⁵. Similar to $\text{Mn}^{2+}$ 

   Mn2+, it has unpaired electrons that can undergo d-d transitions, making it colored.

5. 
   $\text{Co}^{2+}$ 

   Co2+: Has the electronic configuration [Ar] 3d⁷. The presence of unpaired electrons allows d-d transitions, making it colored.

Colorless Ions:

1. 
   $\text{Cu}^{+}$ 

   Cu+: Has the electronic configuration [Ar] 3d¹⁰. The d-orbitals are completely filled, leaving no room for d-d transitions. Therefore, $\text{Cu}^{+}$ 

   Cu+ is colorless.

2. 
   $\text{Sc}^{3+}$ 

   Sc3+: Has the electronic configuration [Ar]. It has no electrons in the d-orbitals, making d-d transitions impossible. Therefore, $\text{Sc}^{3+}$ 

   Sc3+ is colorless.

In summary,

$\text{Ti}^{3+}$

Ti3+,

$\text{V}^{3+}$

V3+,

$\text{Mn}^{2+}$

Mn2+,

$\text{Fe}^{3+}$

Fe3+, and

$\text{Co}^{2+}$

Co2+ are colored due to d-d transitions in incompletely filled d orbitals, while

$\text{Cu}^{+}$

Cu+ and

$\text{Sc}^{3+}$

Sc3+ are colorless due to completely filled or vacant d orbitals, respectively.

Question 19. Compare the stability of the

$+2$

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

Solution:

The oxidation states of the first transition series metals are as follows:

Element	Oxidation States
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
Zn	+2

Analysis of the Stability of

$+2$

+2 Oxidation State:

1. Scandium (Sc): The
   $+2$ 

   $+2$ oxidation state is virtually unknown for Scandium, as Sc $^{3+}$ 

   3+ is the more stable oxidation state due to the stable, empty d-orbital configuration ([Ar]).

2. Titanium (Ti): The
   $+2$ 

   $+2$ state is less stable compared to $+4$ 

   $+4$ due to a higher effective nuclear charge favoring the loss of more electrons to achieve a more stable configuration.

3. Vanadium (V): The
   $+2$ 

   $+2$ state is less stable compared to higher oxidation states like $+5$ 

   $+5$, which has a half-filled d-orbital configuration ([Ar] 3d³) providing extra stability.

4. Chromium (Cr): The
   $+2$ 

   $+2$ state exists but is less stable compared to the $+3$ 

   $+3$ and $+6$ 

   $+6$ states, due to the stability of the half-filled (3d⁵) and fully-filled (3d⁰) configurations, respectively.

5. Manganese (Mn): The
   $+2$ 

   $+2$ state is stable and commonly observed. Mn $^{2+}$ 

   2+ has a half-filled d⁵ configuration ([Ar] 3d⁵), which provides extra stability.

6. Iron (Fe): The
   $+2$ 

   $+2$ state is stable, with Fe $^{2+}$ 

   2+ having a [Ar] 3d⁶ configuration. However, the $+3$ 

   $+3$ state is also stable and frequently observed.

7. Cobalt (Co): The
   $+2$ 

   $+2$ state is stable, with Co $^{2+}$ 

   2+ having a [Ar] 3d⁷ configuration. The $+3$ 

   $+3$ state is less common but also exists.

8. Nickel (Ni): The
   $+2$ 

   $+2$ state is the most stable oxidation state, with Ni $^{2+}$ 

   2+ having a [Ar] 3d⁸ configuration.

9. Copper (Cu): The
   $+2$ 

   $+2$ state is stable, with Cu $^{2+}$ 

   2+ having a [Ar] 3d⁹ configuration. However, the $+1$ 

   $+1$ state is also frequently observed due to the stability of the [Ar] 3d¹⁰ configuration.

10. Zinc (Zn): The
    $+2$ 

    $+2$ state is the only stable oxidation state, with Zn $^{2+}$ 

    2+ having a [Ar] 3d¹⁰ configuration.

In summary, the stability of the

$+2$

+2 oxidation state generally increases across the series from left to right, with some exceptions. This trend is due to the increasing effective nuclear charge and the stabilization of half-filled and fully-filled d-orbital configurations.

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

(i) Electronic configuration

(ii) Atomic and ionic sizes

(iii) Oxidation states

(iv) Chemical reactivity

Solution:

(i) Electronic Configuration

• Lanthanoids: The general electronic configuration is [Xe] 4f¹⁻¹⁴ 5d⁰¹ 6s². The 4f orbitals are more deeply buried and less involved in bonding compared to actinoid 5f orbitals.
• Actinoids: The general electronic configuration is [Rn] 5f¹⁻¹⁴ 6d⁰¹ 7s². The 5f orbitals are not as deeply buried as 4f orbitals and participate more actively in bonding and chemical reactions.

(ii) Atomic and Ionic Sizes

• Lanthanoids: They exhibit lanthanoid contraction, which is a gradual decrease in atomic and ionic radii with increasing atomic number. This contraction is due to the poor shielding effect of the 4f electrons.
• Actinoids: They also exhibit actinoid contraction, which is more pronounced than lanthanoid contraction. The poor shielding effect of the 5f orbitals results in a greater decrease in atomic and ionic sizes as the atomic number increases.

(iii) Oxidation States

• Lanthanoids: The principal oxidation state is +3, but +2 and +4 oxidation states are also observed due to the stability associated with fully-filled and half-filled 4f orbitals.
• Actinoids: They exhibit a wider range of oxidation states due to the comparable energies of the 5f, 6d, and 7s orbitals. The principal oxidation state is also +3, but +4, +5, and +6 oxidation states are commonly found.

(iv) Chemical Reactivity

• Lanthanoids: The reactivity of lanthanides increases with atomic number. The early members react with water to form hydroxides and hydrogen, and their reactivity becomes similar to aluminum (Al) with increasing atomic number.
• Actinoids: Actinoids are highly reactive, especially in their finely divided form. They react with boiling water to form a mixture of oxides and hydrides and combine with most non-metals at moderate temperatures. They are not significantly affected by alkalis but can be slightly affected by acids, such as nitric acid, due to the formation of a protective oxide layer.

Question 21. How would you account for the following:

(i) Of the

$d^4$

d4 species, Cr

$^{2+}$

2+ is strongly reducing while Mn

$^{3+}$

3+ is strongly oxidizing.

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

(iii) The

$d^1$

d1 configuration is very unstable in ions.

Solution:

(i) Cr

$^{2+}$

2+ is strongly reducing while Mn

$^{3+}$

3+ is strongly oxidizing:

• Cr
  $^{2+}$ 

  2+: It has a $d^4$ 

  d4 electronic configuration. When Cr $^{2+}$ 

  2+ acts as a reducing agent, it gets oxidized to Cr $^{3+}$ 

  3+ (which has a $d^3$ 

  d3 configuration). The $d^3$ 

  d3 configuration is more stable due to the half-filled stability of the $t_{2g}$ 

  t2g​ orbitals in an octahedral field. This increased stability of the $d^3$ 

  d3 configuration makes Cr $^{2+}$ 

  2+ a strong reducing agent.

• Mn
  $^{3+}$ 

  3+: It has a $d^4$ 

  d4 configuration. As an oxidizing agent, Mn $^{3+}$ 

  3+ gets reduced to Mn $^{2+}$ 

  2+ (with a $d^5$ 

  d5 configuration). The $d^5$ 

  d5 configuration is particularly stable because it is half-filled, which is energetically favorable. Hence, Mn $^{3+}$ 

  3+ is a strong oxidizing agent due to the drive to achieve the stable $d^5$ 

  d5 configuration.

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

• Cobalt(II) Stability: In aqueous solution, Co
  $^{2+}$ 

  2+ is relatively stable due to the stability of its $[Ar] 3d^7$ 

  [Ar]3d7 electronic configuration.

• Complexing Reagents: In the presence of strong field ligands, the formation of complex ions leads to significant crystal field stabilization energy (CFSE). This stabilization can offset the high ionization energy required to remove an electron and oxidize Co
  $^{2+}$ 

  2+ to Co $^{3+}$ 

  3+. The increased stabilization provided by the complexing agents can make the oxidation to Co $^{3+}$ 

  3+ more favorable.

(iii) The

$d^1$

d1 configuration is very unstable in ions:

• Instability of
  $d^1$ 

  d1 Configuration: Ions with a $d^1$ 

  d1 configuration are unstable because the single electron in the $d$ 

  $d$-orbital is easily lost to achieve a more stable $d^0$ 

  d0 configuration. The energy required to remove this single $d$ 

  $d$-electron is often lower than the lattice or hydration energy, making the $d^1$ 

  d1 configuration less stable. As a result, such ions tend to act as reducing agents to reach the more stable $d^0$ 

  d0 state.

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

Solution:

Disproportionation reactions are a type of redox reaction where a single substance is both oxidized and reduced simultaneously, resulting in the formation of two different products.

Examples:

(a) Decomposition of Hydrogen Peroxide:

$2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2$

2H2​O2​→2H2​O+O2​ In this reaction:

• Oxygen in hydrogen peroxide (
  $\text{H}_2\text{O}_2$ 

  H2​O2​) is oxidized from an oxidation state of -1 to 0 in oxygen gas ( $\text{O}_2$ 

  O2​).

• Simultaneously, oxygen in hydrogen peroxide is reduced from -1 to -2 in water (
  $\text{H}_2\text{O}$ 

  H2​O).

(b) Reaction of Manganate Ion (

$\text{MnO}_4^{2-}$

MnO42−​) in Acidic Medium:

$3 \text{MnO}_4^{2-} + 4 \text{H}^+ \rightarrow 2 \text{MnO}_4^- + \text{MnO}_2 + 2 \text{H}_2\text{O}$

3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O In this reaction:

• Manganese in
  $\text{MnO}_4^{2-}$ 

  MnO42−​ is oxidized from an oxidation state of +6 to +7 in $\text{MnO}_4^-$ 

  MnO4−​.

• Simultaneously, manganese is reduced from +6 to +4 in
  $\text{MnO}_2$ 

  MnO2​.

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

Solution:

In the first transition series, Copper (Cu) exhibits the

$+1$

+1 oxidation state most frequently. This is because in the

$+1$

+1 oxidation state, Copper has the electronic configuration

$[Ar] 3d^{10}$

[Ar]3d10. The completely filled

$3d$

3d-orbitals in Cu

$^{+}$

+ confer extra stability to the ion. This stable

$3d^{10}$

3d10 configuration is why the

$+1$

+1 oxidation state of Copper is so common.

Question 24. Calculate the number of unpaired electrons in the following gaseous ions: Mn

$^{3+}$

3+, Cr

$^{3+}$

3+, V

$^{3+}$

3+, and Ti

$^{3+}$

3+. Which one of these is the most stable in aqueous solution?

Solution:

Gaseous Ions and Number of Unpaired Electrons:

(i) Mn

$^{3+}$

3+: Electronic configuration is

$[Ar] 3d^4$

[Ar]3d4. Number of unpaired electrons = 4.

(ii) Cr

$^{3+}$

3+: Electronic configuration is

$[Ar] 3d^3$

[Ar]3d3. Number of unpaired electrons = 3.

(iii) V

$^{3+}$

3+: Electronic configuration is

$[Ar] 3d^2$

[Ar]3d2. Number of unpaired electrons = 2.

(iv) Ti

$^{3+}$

3+: Electronic configuration is

$[Ar] 3d^1$

[Ar]3d1. Number of unpaired electrons = 1.

Stability in Aqueous Solution:

Cr

$^{3+}$

3+ is the most stable in aqueous solution due to its electronic configuration

$[Ar] 3d^3$

[Ar]3d3, which provides extra stability because of the half-filled

$3d$

3d-orbitals.

Question 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 the highest oxidation state in oxides and fluorides.

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

Solution:

(i) Basicity of Lowest Oxide and Acidity of Highest Oxide:

• The lowest oxide of a transition metal generally has a low oxidation state. In this case, the metal atom has a higher number of valence electrons available for donation. Thus, it tends to act as a base by donating these electrons.Example:

  $\text{MnO}$MnO (manganous oxide) is basic.

• The highest oxide of a transition metal has a high oxidation state, where most of the valence electrons are involved in bonding, and the metal atom exhibits a high effective nuclear charge. This makes it more likely to accept electrons and behave as an acid.Example:

  $\text{MnO}_4^-$MnO4−​ (permanganate ion) is acidic or amphoteric.

(ii) Highest Oxidation State in Oxides and Fluorides:

• Oxygen and fluorine are highly electronegative and small in size. Due to their high electronegativities, they can stabilize transition metals in their highest oxidation states. The strong oxidizing nature of these elements allows the metals to achieve their maximum oxidation states.Examples:
  • 
    $\text{OsF}_6$ 

    OsF6​ (Osmium hexafluoride), where osmium is in the +6 oxidation state.

  • 
    $\text{V}_2\text{O}_5$ 

    V2​O5​ (Vanadium pentoxide), where vanadium is in the +5 oxidation state.

(iii) Highest Oxidation State in Oxoanions:

• In oxoanions, oxygen acts as a strong oxidizing agent due to its high electronegativity and small size. This allows the transition metal to achieve its highest oxidation state, as oxygen effectively stabilizes the metal in this state.Example:
  • 
    $\text{MnO}_4^-$ 

    MnO4−​ (permanganate ion), where manganese is in the +7 oxidation state.

Question 26. Indicate the steps in the preparation of:

(i) K₂Cr₂O₇ from chromite ore.

(ii) KMnO₄ from pyrolusite ore.

Solution:

(i) Preparation of K₂Cr₂O₇ from Chromite Ore:

1. Preparation of Sodium Chromate:
   • Chromite ore (FeCr₂O₄) is heated with sodium carbonate (Na₂CO₃) in the presence of air. This converts the chromium in the ore into sodium chromate (Na₂CrO₄).
   • 
     $\text{FeCr}_2\text{O}_4 + 4\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 2\text{Na}_2\text{CrO}_4 + \text{Fe}_2\text{O}_3 + 4\text{CO}_2$ 

     FeCr2​O4​+4Na2​CO3​+7O2​→2Na2​CrO4​+Fe2​O3​+4CO2​

2. Conversion of Sodium Chromate into Sodium Dichromate:
   • Sodium chromate solution is acidified with sulfuric acid (H₂SO₄) to form sodium dichromate (Na₂Cr₂O₇).
   • 
     $2\text{Na}_2\text{CrO}_4 + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ 

     2Na2​CrO4​+2H2​SO4​→Na2​Cr2​O7​+2Na2​SO4​+2H2​O

3. Conversion of Sodium Dichromate to Potassium Dichromate:
   • Sodium dichromate is treated with potassium chloride (KCl) to obtain potassium dichromate (K₂Cr₂O₇). Potassium dichromate is less soluble in water compared to sodium chloride, so it crystallizes out.
   • 
     $\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}$ 

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

   • The orange-colored potassium dichromate crystals are then separated by filtration.

(ii) Preparation of KMnO₄ from Pyrolusite Ore:

1. Formation of Potassium Manganate:
   • Pyrolusite ore (MnO₂) is fused with potassium hydroxide (KOH) in the presence of an oxidizing agent like potassium nitrate (KNO₃) or potassium perchlorate (KClO₄) to form potassium manganate (K₂MnO₄).
   • 
     $2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}$ 

     2MnO2​+4KOH+O2​→2K2​MnO4​+2H2​O

2. Oxidation to Potassium Permanganate:
   • The green potassium manganate solution is then oxidized to potassium permanganate (KMnO₄) either by electrolytic oxidation or by passing chlorine (Cl₂) or ozone (O₃) through the solution.
   • Electrolytic Oxidation:
     • At the anode, manganate ions (MnO₄²⁻) are oxidized to permanganate ions (MnO₄⁻).
   • Oxidation by Chlorine/Ozone:
     • The green solution of potassium manganate is treated with chlorine or ozone, which oxidizes MnO₄²⁻ to MnO₄⁻.
     • 
       $2\text{K}_2\text{MnO}_4 + \text{Cl}_2 \rightarrow 2\text{KMnO}_4 + 2\text{KCl}$ 

       2K2​MnO4​+Cl2​→2KMnO4​+2KCl

     • 
       $2\text{K}_2\text{MnO}_4 + \text{O}_3 \rightarrow 2\text{KMnO}_4 + \text{O}_2$ 

       2K2​MnO4​+O3​→2KMnO4​+O2​

   • The purple-colored potassium permanganate crystals are then obtained by evaporating the water and can be purified further if needed.

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

Solution:

An alloy is a solid solution of two or more elements, where at least one of them is a metal. Alloys are typically characterized by their metallic matrix and can be either a partial or complete solid solution. The properties of alloys often differ significantly from those of the individual components, providing enhanced physical, chemical, or mechanical properties.

An important alloy containing lanthanoids is Mischmetal. Mischmetal typically contains 94-95% lanthanoids, including elements like cerium, lanthanum, and neodymium, as well as small amounts of iron (about 5%) and traces of other elements such as sulfur (S), carbon (C), silicon (Si), calcium (Ca), and aluminum (Al).

Uses of Mischmetal:

• In Lighters and Gas Stoves: Mischmetal is used in the production of flint for cigarette lighters and gas lighters due to its ability to produce sparks when struck.
• In Flame Throwing Tanks: It is used in the flame-throwing tanks where its properties assist in creating an effective flame.
• In Tracer Bullets and Shells: Mischmetal is also used in tracer bullets and artillery shells to improve performance and visibility.

Question 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:

Inner transition elements are those elements in which the last electron enters the f-orbital. These elements are also known as f-block elements and include:

• Lanthanides: Elements where the 4f orbitals are progressively filled.
• Actinides: Elements where the 5f orbitals are progressively filled.

Among the given atomic numbers, the atomic numbers corresponding to inner transition elements are:

• 59: This is the atomic number of Neodymium (Nd), which is a lanthanide.
• 95: This is the atomic number of Americium (Am), which is an actinide.
• 102: This is the atomic number of Nobelium (No), which is an actinide.

The atomic numbers 29 (Copper), 74 (Tungsten), and 104 (Rutherfordium) do not correspond to inner transition elements.

Question 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.

Solution:

The chemistry of actinoid elements is more complex compared to lanthanoids due to the following reasons:

1. Energy Levels and Oxidation States:
   • Lanthanoids exhibit a limited range of oxidation states, primarily +2, +3, and occasionally +4. The +3 oxidation state is the most stable and common due to the relatively large energy difference between the 4f, 5d, and 6s orbitals, which restricts the availability of electrons for oxidation.
   • Actinoids, on the other hand, exhibit a wider range of oxidation states, from +3 to +7. This is because the energy difference between the 5f, 6d, and 7s orbitals is smaller compared to the lanthanoids, allowing for a greater number of accessible oxidation states. For example:
     • Uranium (U) shows oxidation states of +3, +4, +5, and +6.
     • Plutonium (Pu) shows oxidation states of +3, +4, +5, and +6.
     • Neptunium (Np) displays oxidation states of +3, +4, +5, and +7.
2. Chemical Behavior:
   • The larger number of oxidation states and the ability to form multiple oxidation states make the chemistry of actinoids more diverse and complex. This contrasts with lanthanoids, which typically have more predictable and consistent chemical behavior due to fewer oxidation states.

In summary, the greater number of accessible oxidation states and the smaller energy gap between the 5f, 6d, and 7s orbitals in actinoids contribute to their more complex and less predictable chemistry compared to the lanthanoids.

Question 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.

Solution:

The last element in the actinoid series is lawrencium (Lr), with atomic number 103.

Electronic Configuration: The electronic configuration of lawrencium is

$[Rn] 5f^{14} 6d^1 7s^2$

[Rn]5f146d17s2.

Possible Oxidation States:

• The most common oxidation state of lawrencium is +3. This is because, after losing three electrons, it achieves a stable f
  $^14$ 

  14 configuration, similar to the stable configurations seen in other actinoids.

• Although lawrencium can theoretically exhibit other oxidation states, the +3 state is the most stable and frequently observed due to the relatively high energy of the 6d and 7s electrons compared to the 5f electrons.

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

Solution:

1. Deriving the Electronic Configuration of Ce³⁺:

• Atomic Number of Cerium (Ce): 58
• Ground-State Electronic Configuration of Ce:
  $[Xe] 4f^1 5d^1 6s^2$ 

  [Xe]4f15d16s2

To derive the electronic configuration of Ce³⁺:

1. Remove three electrons from the neutral atom. The electrons are removed first from the 6s orbital and then from the 5d orbital (as they are the outermost electrons).
2. The configuration of Ce³⁺ is: 

   $$\text{Ce} \rightarrow [Xe] 4f^1 5d^1 6s^2$$Ce→[Xe]4f15d16s2Removing three electrons:

    

   $$[Xe] 4f^1 5d^1 6s^2 \rightarrow [Xe] 4f^1$$[Xe]4f15d16s2→[Xe]4f1Therefore, the electronic configuration of Ce³⁺ is:

    

   $$[Xe] 4f^1$$[Xe]4f1

2. Calculating the Magnetic Moment:

• Number of Unpaired Electrons (n): 1 (since Ce³⁺ has a single electron in the 4f orbital)

The magnetic moment (

$\mu$

μ) is calculated using the formula:

 

$$\mu =\sqrt{n(n+2)}\text{B.M.}$$

 

For

$n = 1$

n=1:

 

$$\mu =\sqrt{1(1+2)}=\sqrt{3}\approx 1.732\text{B.M.}$$

 

So, the magnetic moment of Ce³⁺ is approximately 1.732 Bohr Magneton (B.M.).

Question 32. Name the members of the lanthanoid series which exhibit

$+4$

+4 oxidation states and those which exhibit

$+2$

+2 oxidation states. Try to correlate this type of behavior with the electronic configurations of these elements.

Solution:

1. Lanthanoids Exhibiting

$+4$

+4 Oxidation State:

• Cerium (Ce,
  $Z = 58$ 

  $Z=58$)

• Praseodymium (Pr,
  $Z = 59$ 

  $Z=59$)

• Neodymium (Nd,
  $Z = 60$ 

  $Z=60$)

• Terbium (Tb,
  $Z = 65$ 

  $Z=65$)

• Dysprosium (Dy,
  $Z = 66$ 

  $Z=66$)

Reason: These elements exhibit the

$+4$

+4 oxidation state by losing four electrons, which can lead to stable electronic configurations. For example:

• Ce (Ce³⁺):
  $[Xe] 4f^1$ 

  [Xe]4f1. Losing one more electron (Ce⁴⁺) achieves $[Xe] 4f^0$ 

  [Xe]4f0 (a more stable configuration due to empty 4f orbital).

• Pr (Pr⁴⁺):
  $[Xe] 4f^3$ 

  [Xe]4f3. Losing additional electrons (Pr⁴⁺) results in $[Xe] 4f^2$ 

  [Xe]4f2 or $[Xe] 4f^0$ 

  [Xe]4f0.

• Nd (Nd⁴⁺):
  $[Xe] 4f^4$ 

  [Xe]4f4. Losing electrons yields $[Xe] 4f^3$ 

  [Xe]4f3 or $[Xe] 4f^0$ 

  [Xe]4f0.

2. Lanthanoids Exhibiting

$+2$

+2 Oxidation State:

• Neodymium (Nd,
  $Z = 60$ 

  $Z=60$)

• Samarium (Sm,
  $Z = 62$ 

  $Z=62$)

• Europium (Eu,
  $Z = 63$ 

  $Z=63$)

• Thulium (Tm,
  $Z = 69$ 

  $Z=69$)

• Ytterbium (Yb,
  $Z = 70$ 

  $Z=70$)

Reason: These elements exhibit the

$+2$

+2 oxidation state due to the stability associated with half-filled and fully filled 4f orbitals:

• Eu (Eu²⁺):
  $[Xe] 4f^7$ 

  [Xe]4f7. The half-filled $4f^7$ 

  4f7 configuration is stable.

• Yb (Yb²⁺):
  $[Xe] 4f^14$ 

  [Xe]4f14. The fully filled $4f^{14}$ 

  4f14 configuration is highly stable.

Correlation with Electronic Configurations:

• Stability of
  $+4$ 

  $+4$ State: This oxidation state is stabilized by achieving electronic configurations close to $4f^0$ 

  4f0 (empty 4f orbitals) or $4f^7$ 

  4f7 (which allows for a stable distribution of electrons).

• Stability of
  $+2$ 

  $+2$ State: The $+2$ 

  $+2$ oxidation state is stabilized by achieving half-filled $4f^7$ 

  4f7 or fully filled $4f^{14}$ 

  4f14 configurations, which offer extra stability due to exchange energy and symmetrical distribution of electrons.

The stability of these oxidation states is strongly influenced by the electronic configuration of the 4f orbitals, as elements tend to adopt oxidation states that bring their electron configurations closer to stable arrangements.

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

(i) Electronic configuration

(ii) Oxidation states

(iii) Chemical reactivity

Solution:

Electronic configuration:

• Lanthanoids: [Xe] 4f¹⁻¹⁴ 5d⁰¹ 6s²
• Actinoids: [Rn] 5f¹⁻¹⁴ 6d⁰¹ 7s²

Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

Oxidation states:

• Lanthanoids: Principal oxidation state is +3. Other oxidation states include +2 and +4 due to the stability of fully-filled and half-filled orbitals.
• Actinoids: Exhibit a greater range of oxidation states due to comparable energies of 5f, 6d, and 7s levels. Principal oxidation state is also +3, but a wider range of oxidation states is observed compared to lanthanoids.

Chemical reactivity:

• Lanthanoids: Earlier members are more reactive, comparable to Ca. Reactivity increases with atomic number, approaching that of Al.
• Actinoids: Highly reactive, especially when finely divided. React with boiling water to form oxides and hydrides, combine with most non-metals at moderate temperatures, and are slightly affected by acids due to the formation of a protective oxide layer.

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

Sol: The electronic configurations of the elements with the given atomic numbers are shown below:

 

$$\begin{array}{|c|c|c|} \hline \textbf{Atomic Number} & \textbf{Element} & \textbf{Electronic Configuration} \\ \hline 61 & \text{Promethium (Pm)} & [Xe] \, 4f^5 \, 6s^2 \\ 91 & \text{Protactinium (Pa)} & [Rn] \, 5f^2 \, 6d^1 \, 7s^2 \\ 101 & \text{Mendelevium (Md)} & [Rn] \, 5f^{13} \, 6d^0 \, 7s^2 \\ 109 & \text{Meitnerium (Mt)} & [Rn] \, 5f^{14} \, 6d^7 \, 7s^2 \\ \hline \end{array}$$

​​​​

Question 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:

• A. Electronic configuration
• B. Oxidation states
• C. Ionization enthalpies
• D. Atomic size

Solution

A) Electronic Configuration

Transition metals are characterized by the filling of

$d$

d-orbitals. The general electronic configuration for

$d$

d-block elements is

$(n-1)d^{1-10}ns^{1-2}$

(n−1)d1−10ns1−2. In the first, second, and third transition series, the

$3d$

3d,

$4d$

4d, and

$5d$

5d orbitals are respectively filled. The elements in the same vertical column generally have similar electronic configurations.

• First Transition Series (3d series): In this series, the last electron enters the
  $3d$ 

  $3d$ subshell. Two elements show unusual electronic configurations due to highly stable half-filled and completely filled orbitals:

  • Copper (Cu,
    $Z = 29$ 

    $Z=29$): [Ar] $3d^{10}4s^1$ 

    3d104s1

  • Chromium (Cr,
    $Z = 24$ 

    $Z=24$): [Ar] $3d^5 4s^1$ 

    3d54s1

• Second Transition Series (4d series): In this series, the last electron enters the
  $4d$ 

  $4d$ subshell. Some elements show unusual electronic configurations:

  • Silver (Ag,
    $Z = 47$ 

    $Z=47$): [Kr] $4d^{10}5s^1$ 

    4d105s1

  • Palladium (Pd,
    $Z = 46$ 

    $Z=46$): [Kr] $4d^{10}5s^0$ 

    4d105s0

  • Rhodium (Rh,
    $Z = 45$ 

    $Z=45$): [Kr] $4d^8 5s^1$ 

    4d85s1

  • Ruthenium (Ru,
    $Z = 44$ 

    $Z=44$): [Kr] $4d^7 5s^1$ 

    4d75s1

  • Molybdenum (Mo,
    $Z = 42$ 

    $Z=42$): [Kr] $4d^5 5s^1$ 

    4d55s1

• Third Transition Series (5d series): In this series, the last electron enters the
  $5d$ 

  $5d$ subshell. Three elements show unusual electronic configurations:

  • Gold (Au,
    $Z = 79$ 

    $Z=79$): [Xe] $5d^{10}6s^1$ 

    5d106s1

  • Platinum (Pt,
    $Z = 78$ 

    $Z=78$): [Xe] $5d^9 6s^1$ 

    5d96s1

  • Tungsten (W,
    $Z = 74$ 

    $Z=74$): [Xe] $5d^4 6s^2$ 

    5d46s2

B) Oxidation States

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,

$+2$

+2 and

$+3$

+3 oxidation states are quite stable for all elements present in the first transition series, with Cr and Mn showing special oxidation states.

• First Transition Series: All metals form stable compounds in the
  $+2$ 

  $+2$ and $+3$ 

  $+3$ oxidation states.

• Second and Third Transition Series: The stability of the
  $+2$ 

  $+2$ and $+3$ 

  $+3$ oxidation states decreases, and higher oxidation states become more important. The maximum oxidation state in a group increases from the first transition series to the third transition series. For example:

  • Group 8: Iron (Fe) shows
    $+2$ 

    $+2$ and $+3$ 

    $+3$ oxidation states, Ruthenium (Ru) shows $+4$ 

    $+4$ and $+6$ 

    $+6$, and Osmium (Os) shows $+8$ 

    $+8$ oxidation state.

C) Ionization Enthalpies

Ionization enthalpy is the amount of energy required to remove an electron from an isolated gaseous atom. Factors affecting ionization enthalpies include:

1. Penetration Effect: The relative density of electrons near the nucleus affects ionization energy.
2. Shielding Effect: Inner electrons shield outer electrons from the full charge of the nucleus, affecting the ionization energy.
3. Electronic Configuration: Half-filled and fully filled orbitals are stable, requiring more energy to remove electrons.

• In each series, the first ionization enthalpy generally increases from left to right, with exceptions. The first ionization enthalpies of the third transition series are higher than those of the second and first series due to the poor shielding effect of
  $4f$ 

  $4f$ electrons.

D) Atomic Size

Atomic size decreases across a period due to increasing effective nuclear charge. However, atomic sizes in the second transition series are generally larger than those in the first series due to the addition of an extra shell. In the third series, atomic sizes are similar to the second series due to lanthanoid contraction, which results from poor shielding by

$4f$

4f electrons. Trends in atomic radii of transition elements show a general decrease from left to right, with the lanthanoid contraction accounting for similar sizes in the second and third series.

Question 36. Write down the number of $3d$ electrons in each of the following ions: $\text{Ti}^{2+}$, $\text{V}^{2+}$, $\text{Cr}^{3+}$, $\text{Mn}^{2+}$, $\text{Fe}^{2+}$, $\text{Fe}^{3+}$, $\text{Co}^{2+}$, $\text{Ni}^{2+}$, and $\text{Cu}^{2+}$. Indicate how you would expect the five $3d$ orbitals to be occupied for these hydrated ions (octahedral).

Solution:

The number of $3d$ electrons and the occupancy of $3d$ orbitals for various transition metal ions are as follows:

• $\text{Ti}^{2+}$: $3d^2$ — $t_{2g}^2 e_g^0$
• $\text{V}^{2+}$: $3d^3$ — $t_{2g}^3 e_g^0$
• $\text{Cr}^{3+}$: $3d^3$ — $t_{2g}^3 e_g^0$
• $\text{Mn}^{2+}$: $3d^5$ — $t_{2g}^3 e_g^2$
• $\text{Fe}^{2+}$: $3d^6$ — $t_{2g}^4 e_g^2$
• $\text{Fe}^{3+}$: $3d^5$ — $t_{2g}^3 e_g^2$
• $\text{Co}^{2+}$: $3d^7$ — $t_{2g}^5 e_g^2$
• $\text{Ni}^{2+}$: $3d^8$ — $t_{2g}^6 e_g^2$
• $\text{Cu}^{2+}$: $3d^9$ — $t_{2g}^6 e_g^3$

The notation $t_{2g}$ and $e_g$ refers to the splitting of $d$ orbitals in an octahedral crystal field, where $t_{2g}$ represents the lower-energy set of three orbitals and $e_g$ represents the higher-energy set of two orbitals.

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

Solution:

The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways:

(i) Atomic Size: The atomic sizes of the elements in the first transition series are smaller than those of the heavier elements (elements of the second and third 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.

(ii) Oxidation States: The

$+2$

+2 and

$+3$

+3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.

(iii) Enthalpies of Atomisation: 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.

(iv) Melting and Boiling Points: 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).

(v) Complex Formation: 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.

Question 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:

Magnetic moment (

$\mu$

μ) is given by

$\mu = \sqrt{n(n + 2)}$

μ=n(n+2)​.

For different values of

$n$

n:

• For
  $n = 1$ 

  $n=1$, $\mu = \sqrt{1(1 + 2)} = \sqrt{3} = 1.73$ 

  μ=1(1+2)​=3​=1.73

• For
  $n = 2$ 

  $n=2$, $\mu = \sqrt{2(2 + 2)} = \sqrt{8} = 2.83$ 

  μ=2(2+2)​=8​=2.83

• For
  $n = 3$ 

  $n=3$, $\mu = \sqrt{3(3 + 2)} = \sqrt{15} = 3.87$ 

  μ=3(3+2)​=15​=3.87

• For
  $n = 4$ 

  $n=4$, $\mu = \sqrt{4(4 + 2)} = \sqrt{24} = 4.90$ 

  μ=4(4+2)​=24​=4.90

• For
  $n = 5$ 

  $n=5$, $\mu = \sqrt{5(5 + 2)} = \sqrt{35} = 5.92$ 

  μ=5(5+2)​=35​=5.92

(i) K₄[Mn(CN)₆]

• Given
  $\mu = 2.2$ 

  $\mu =2.2$

• The closest value is
  $n = 1$ 

  $n=1$ ( $\mu \approx 1.73$ 

  $\mu \approx 1.73$ )

• Mn is in the +2 oxidation state, with 5 d-electrons.
• CN⁻ is a strong field ligand that causes pairing of electrons.

(ii) [Fe(H₂O)₆]²⁺

• Given
  $\mu = 5.3$ 

  $\mu =5.3$

• The closest value is
  $n = 4$ 

  $n=4$ ( $\mu \approx 4.90$ 

  $\mu \approx 4.90$ )

• Fe is in the +2 oxidation state, with 6 d-electrons.
• H₂O is a weak field ligand and does not cause pairing of electrons.

(iii) K₂[MnCl₄]

• Given
  $\mu = 5.9$ 

  $\mu =5.9$

• The closest value is
  $n = 5$ 

  $n=5$ ( $\mu \approx 5.92$ 

  $\mu \approx 5.92$ )

• Mn is in the +2 oxidation state, with 5 d-electrons.
• Cl⁻ is a weak field ligand and does not cause pairing of electrons.

Question 39. Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?

Solution:

Ag has a completely filled 4d orbital (4d¹⁰ 5s¹) in its ground state. Silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orbital. However, in the +2 oxidation state, an electron is removed from the d-orbital, resulting in an incomplete d-orbital (4d⁹). Hence, silver is classified as a transition element.

Question 40. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol⁻¹. Why?

Solution:

The extent of metallic bonding an element undergoes determines the enthalpy of atomization. The more extensive the metallic bonding, the higher the enthalpy of atomization. In all transition metals (except Zn, with electronic configuration 3d¹⁰ 4s²), there are unpaired electrons contributing to stronger metallic bonding. Due to the absence of unpaired electrons in Zn, the inter-atomic electronic bonding is weaker, resulting in the lowest enthalpy of atomization.

Question 41. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Solution:

Mn (Z = 25) = 3d⁵ 4s²

Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.

Question 42. The $E^\circ (\text{M}^{2+}/\text{M})$ value for copper is positive (+0.34 V). What is the possible reason for this?

Solution:

Copper has a high atomization energy ($\Delta_a H^\circ$) and low hydration energy ($\Delta_{hyd} H^\circ$), resulting in a positive standard electrode potential ($E^\circ$) for the reduction of $\text{Cu}^{2+}$ to $\text{Cu}$. The positive $E^\circ$ value indicates that the reduction of copper ions to metallic copper is not highly favorable under standard conditions.

Question 43. How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?

Solution:

Ionization enthalpies generally increase in the series due to the continuous filling of the inner d-orbitals. The irregular variations can be attributed to the extra stability associated with certain electron configurations, such as $d^0$, $d^5$, and $d^{10}$. These stable configurations result in higher ionization enthalpies.

First Ionization Energy: For example, Chromium (Cr) has a lower first ionization energy because, after losing one electron, it attains the stable $3d^5$ configuration. In contrast, Zinc (Zn) has a higher first ionization energy as removing an electron from the stable and fully-filled $3d^{10}4s^2$ orbitals requires more energy.

Second Ionization Energy: The second ionization energies are higher than the first because removing a second electron from a positively charged ion is more difficult. Elements like Chromium (Cr) and Copper (Cu) exhibit exceptionally high second ionization energies because, after losing the first electron, they attain stable configurations (Cr$^+$: $3d^5$ and Cu$^+$: $3d^{10}$). Removing an additional electron from these stable configurations requires significantly more energy.

Question 44. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Solution:

Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.

Question 45. Which is a stronger reducing agent: $\text{Cr}^{2+}$ or $\text{Fe}^{2+}$ and why?

Solution:

The following reactions are involved when $\text{Cr}^{2+}$ and $\text{Fe}^{2+}$ act as reducing agents:

$$\text{Cr}^{2+} \rightarrow \text{Cr}^{3+}$$ $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$$

The standard electrode potential for $\text{Cr}^{2+}/\text{Cr}^{3+}$ is −0.41 V, and for $\text{Fe}^{2+}/\text{Fe}^{3+}$ is +0.77 V. This indicates that $\text{Cr}^{2+}$ can be easily oxidized to $\text{Cr}^{3+}$, while $\text{Fe}^{2+}$ does not get oxidized to $\text{Fe}^{3+}$ as readily. Therefore, $\text{Cr}^{2+}$ is a better reducing agent than $\text{Fe}^{2+}$.

Question 46. Calculate the ‘spin only’ magnetic moment of

$\text{M}^{2+}(\text{aq})$

M2+(aq) ion (

$Z = 27$

Z=27).

Solution:

Number of unpaired electrons:

Given: Atomic number (

$Z$

Z) = 27

The valence electronic configuration of cobalt (

$\text{Co}$

Co) is

$3d^7, 4s^2.$

3d7,4s2.

For the

$\text{M}^{2+}(\text{aq})$

M2+(aq) ion, it loses two electrons. Hence, the valence electronic configuration becomes

$3d^7$

3d7. The number of unpaired electrons is 3.

Spin only magnetic moment:

We know that

$\mu = \sqrt{n(n+2)}$

μ=n(n+2)​ where

$n = \text{number of unpaired electrons}$

n=number of unpaired electrons.

Putting

$n = 3$

n=3, we get:

$\mu = \sqrt{3(3+2)} = 3.87 \, \text{BM}$

μ=3(3+2)​=3.87BM

Final answer:

$\mu = 3.87 \, \text{BM}$

μ=3.87BM

Question 47. Explain why $\text{Cu}^+$ ion is not stable in aqueous solutions.

Solution:

In an aqueous medium, $\text{Cu}^{2+}$ is more stable than $\text{Cu}^+$. Although removing one electron from $\text{Cu}^+$ to form $\text{Cu}^{2+}$ requires energy, the high hydration energy of $\text{Cu}^{2+}$ compensates for this. As a result, $\text{Cu}^+$ ion is unstable in aqueous solution and disproportionates to form $\text{Cu}^{2+}$ and $\text{Cu}$.

Question 48. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Solution:

In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect compared to the 4f orbitals in lanthanoids. As a result, the effective nuclear charge experienced by electrons in the valence shells of actinoids is much greater than that experienced by lanthanoids. Consequently, the size contraction in actinoids is greater compared to that in lanthanoids.