Q: Q11. Complete the following equations:\n (i) Cr2O7^2- + 2OH- →\n (ii) MnO4^2- + 4H+ →\n (iii) Cr2O7^2- + 14H+ + 6Fe^2+ →\n (iv) MnO4- + 8H+ + 5Fe^2+ →\n (v) MnO4- + 8H+ +

Here are the completed and balanced chemical equations:\n\n(i) **Cr2O7^2- + 2OH- → 2CrO4^2- + H2O**\n * (Dichromate ion in basic medium converts to chromate ion)\n\n(ii) **3MnO4^2- + 4H+ → 2MnO4- + MnO2 + 2H2O**\n * (Manganate ion disproportionates in acidic medium)\n\n(iii) **Cr2O7^2- + 14H+ + 6Fe^2+ → 2Cr3+ + 6Fe3+ + 7H2O**\n * (Dichromate ion oxidizes Fe2+ to Fe3+ in acidic medium)\n\n(iv) **MnO4- + 8H+ + 5Fe^2+ → Mn2+ + 5Fe3+ + 4H2O**\n * (Permanganate ion oxidizes Fe2+ to Fe3+ in acidic medium)\n\n(v) **2MnO4- + 16H+ + 5C2O4^2- → 2Mn2+ + 10CO2 + 8H2O**\n * (Permanganate ion oxidizes oxalate ion to CO2 in acidic medium)\n\n(vi) **2MnO4- + 16H+ + 10I- → 2Mn2+ + 5I2 + 8H2O**\n * (Permanganate ion oxidizes iodide ion to iodine in acidic medium)

Question 1

Q1. Write down the electronic configuration of:
(i) Cr³⁺
(ii) Pm³⁺
(iii) Cu⁺
(iv) Ce⁴⁺
(v) Co²⁺
(vi) Lu²⁺
(vii) Mn²⁺
(viii) Th⁴⁺

Accepted Answer

The electronic configurations are determined by first writing the ground state configuration of the neutral atom and then removing electrons from the outermost shell (ns first, then (n-1)d) based on the charge.

(i) **Cr³⁺**
   Neutral Cr (Z=24): [Ar] 3d⁵ 4s¹ (exception to Aufbau principle)
   Cr³⁺: Remove 1 electron from 4s and 2 electrons from 3d.
   Electronic configuration: **[Ar] 3d³**

(ii) **Pm³⁺**
    Neutral Pm (Promethium, Z=61): [Xe] 4f⁵ 6s²
    Pm³⁺: Remove 2 electrons from 6s and 1 electron from 4f.
    Electronic configuration: **[Xe] 4f⁴**

(iii) **Cu⁺**
     Neutral Cu (Z=29): [Ar] 3d¹⁰ 4s¹ (exception to Aufbau principle)
     Cu⁺: Remove 1 electron from 4s.
     Electronic configuration: **[Ar] 3d¹⁰**

(iv) **Ce⁴⁺**
     Neutral Ce (Cerium, Z=58): [Xe] 4f¹ 5d¹ 6s²
     Ce⁴⁺: Remove 2 electrons from 6s, 1 electron from 5d, and 1 electron from 4f.
     Electronic configuration: **[Xe]** (achieves stable noble gas configuration)

(v) **Co²⁺**
    Neutral Co (Z=27): [Ar] 3d⁷ 4s²
    Co²⁺: Remove 2 electrons from 4s.
    Electronic configuration: **[Ar] 3d⁷**

(vi) **Lu²⁺**
     Neutral Lu (Lutetium, Z=71): [Xe] 4f¹⁴ 5d¹ 6s²
     Lu²⁺: Remove 2 electrons from 6s.
     Electronic configuration: **[Xe] 4f¹⁴ 5d¹**

(vii) **Mn²⁺**
      Neutral Mn (Z=25): [Ar] 3d⁵ 4s²
      Mn²⁺: Remove 2 electrons from 4s.
      Electronic configuration: **[Ar] 3d⁵**

(viii) **Th⁴⁺**
       Neutral Th (Thorium, Z=90): [Rn] 6d² 7s²
       Th⁴⁺: Remove 2 electrons from 7s and 2 electrons from 6d.
       Electronic configuration: **[Rn]** (achieves stable noble gas configuration)

Question 2

Q2. Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?

Accepted Answer

The stability of oxidation states in transition elements is often related to achieving stable electronic configurations, such as half-filled (d⁵) or completely filled (d¹⁰) d-orbitals.

1.  **Electronic Configuration of Mn²⁺ and Fe²⁺:**
    *   Neutral Mn (Z=25): [Ar] 3d⁵ 4s²
    *   **Mn²⁺**: [Ar] 3d⁵ (after losing two 4s electrons)
    *   Neutral Fe (Z=26): [Ar] 3d⁶ 4s²
    *   **Fe²⁺**: [Ar] 3d⁶ (after losing two 4s electrons)

2.  **Oxidation to +3 State:**
    *   Oxidation of Mn²⁺ to Mn³⁺ involves the loss of one electron from Mn²⁺:
        Mn²⁺ ([Ar] 3d⁵) → Mn³⁺ ([Ar] 3d⁴) + e⁻
    *   Oxidation of Fe²⁺ to Fe³⁺ involves the loss of one electron from Fe²⁺:
        Fe²⁺ ([Ar] 3d⁶) → Fe³⁺ ([Ar] 3d⁵) + e⁻

3.  **Stability Comparison:**
    *   **Mn²⁺** has a stable half-filled 3d⁵ electronic configuration. This configuration has maximum exchange energy and symmetrical distribution of electrons, making it exceptionally stable. Therefore, Mn²⁺ strongly resists losing another electron to form Mn³⁺ (which would have a less stable 3d⁴ configuration).
    *   **Fe²⁺** has a 3d⁶ electronic configuration. By losing one electron, it can achieve the more stable half-filled 3d⁵ configuration (Fe³⁺). This tendency makes Fe²⁺ readily oxidizable to Fe³⁺.

**Conclusion:** Because Mn²⁺ already possesses a highly stable half-filled 3d⁵ configuration, it is more stable and less prone to oxidation to the +3 state compared to Fe²⁺, which can achieve the stable half-filled 3d⁵ configuration by getting oxidized to Fe³⁺.

Question 3

Q3. What is the basic difference between the electronic configuration of transition and inner transition elements?

Accepted Answer

The basic difference between the electronic configuration of transition and inner transition elements lies in the orbital being filled.

**Transition Elements (d-block elements):**
These elements are characterized by the filling of the (n-1)d orbitals. Their general electronic configuration is (n-1)d¹⁻¹⁰ ns¹⁻².

**Inner Transition Elements (f-block elements):**
These elements are characterized by the filling of the (n-2)f orbitals. Their general electronic configuration is (n-2)f¹⁻¹⁴ (n-1)d⁰⁻¹ ns².

**Summary of Difference:**
*   **Transition elements:** Last electron enters the penultimate (n-1)d subshell.
*   **Inner transition elements:** Last electron enters the anti-penultimate (n-2)f subshell.

Question 4

Q4. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

Accepted Answer

The variation of ionisation enthalpies (first and second) in the first series of transition elements (Sc to Zn) is irregular due to a complex interplay of factors such as electronic configuration, nuclear charge, shielding effect, and the stability associated with half-filled and fully-filled orbitals.

**General Trend:**
Generally, across a period, as the atomic number increases, the nuclear charge increases, and the atomic size decreases. This leads to an expected increase in ionisation enthalpy from left to right.

**Irregularities and Explanations:**

**1. First Ionisation Enthalpy (ΔᵢH₁):**
While there is a general increasing trend, several irregularities are observed:
*   **Chromium (Cr):** The ΔᵢH₁ of Cr (3d⁵ 4s¹) is lower than expected. This is because after removing one electron from the 4s orbital, it achieves a stable half-filled 3d⁵ configuration (Cr⁺: 3d⁵). This stability makes it relatively easier to remove the first electron. *   **Manganese (Mn):** The ΔᵢH₁ of Mn (3d⁵ 4s²) is higher than expected. This is due to the stable half-filled 3d⁵ configuration. Removing an electron from the 4s orbital requires more energy as the resulting Mn⁺ ion (3d⁵ 4s¹) would still maintain the stable 3d⁵ configuration, but the increased nuclear charge and effective nuclear charge make it harder to remove the electron. *   **Iron (Fe):** The ΔᵢH₁ of Fe (3d⁶ 4s²) is lower than Mn. After removing one 4s electron, Fe⁺ (3d⁶ 4s¹) has a 3d⁶ configuration, which is less stable than 3d⁵. The pairing energy in 3d⁶ also contributes to a relatively easier removal of the second 4s electron. *   **Nickel (Ni):** The ΔᵢH₁ of Ni (3d⁸ 4s²) is higher than Fe. The increasing nuclear charge and poor shielding by d-electrons lead to a higher effective nuclear charge, making it harder to remove the first electron. *   **Copper (Cu):** The ΔᵢH₁ of Cu (3d¹⁰ 4s¹) is lower than Ni. This is because removing the single 4s electron leads to a stable fully-filled 3d¹⁰ configuration (Cu⁺: 3d¹⁰). *   **Zinc (Zn):** The ΔᵢH₁ of Zn (3d¹⁰ 4s²) is significantly higher. This is due to the very stable fully-filled 3d¹⁰ and 4s² configurations. A large amount of energy is required to remove an electron from this stable configuration.

**2. Second Ionisation Enthalpy (ΔᵢH₂):**
Similar irregularities are observed for ΔᵢH₂:
*   **Chromium (Cr):** The ΔᵢH₂ of Cr is very high. After removing the first electron, Cr⁺ has a stable 3d⁵ configuration. Removing a second electron from this stable half-filled 3d⁵ orbital requires a significantly large amount of energy. *   **Manganese (Mn):** The ΔᵢH₂ of Mn is relatively lower than Cr. After removing the first electron, Mn⁺ has a 3d⁵ 4s¹ configuration. Removing the second electron from the 4s orbital is relatively easier than removing an electron from the stable 3d⁵ configuration of Cr⁺. *   **Copper (Cu):** The ΔᵢH₂ of Cu is very high. After removing the first electron, Cu⁺ has a very stable fully-filled 3d¹⁰ configuration. Removing a second electron from this stable 3d¹⁰ orbital requires a very large amount of energy. *   **Zinc (Zn):** The ΔᵢH₂ of Zn is also high, but lower than Cu. After removing the first electron, Zn⁺ has a 3d¹⁰ 4s¹ configuration. Removing the second electron from the 4s orbital is relatively easier than removing an electron from the stable 3d¹⁰ configuration of Cu⁺.

**Conclusion:**
The irregular variations in ionisation enthalpies are primarily attributed to the varying stability of d-orbital configurations (especially half-filled 3d⁵ and fully-filled 3d¹⁰) and the relative ease or difficulty of removing electrons from 4s or 3d orbitals, influenced by effective nuclear charge and shielding effects.

Question 5

Q5. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Accepted Answer

The highest oxidation state of a metal is predominantly exhibited in its oxide or fluoride compounds due to the following reasons:

1.  **High Electronegativity:** Oxygen and fluorine are the most electronegative elements in the periodic table. Their high electronegativity allows them to draw electron density effectively from the metal, leading to the metal exhibiting its highest possible oxidation state. 2.  **Small Size:** Both oxygen and fluorine atoms are very small in size. This small size allows a large number of oxygen or fluorine atoms to surround a central metal atom without significant steric hindrance. This enables the formation of multiple bonds or coordination with several electronegative atoms, facilitating the removal of multiple electrons from the metal to achieve its highest oxidation state. For example, in Mn₂O₇, manganese is in the +7 oxidation state, and in OsF₆, osmium is in the +6 oxidation state. 3.  **Ability to Form Multiple Bonds (Oxygen):** Oxygen has the ability to form multiple bonds (e.g., double bonds) with metals, which helps in achieving higher oxidation states. 4.  **High Lattice Energy/Bond Energy:** The compounds formed with oxygen and fluorine (oxides and fluorides) are generally very stable due to high lattice energies (for ionic compounds) or high bond energies (for covalent compounds). The formation of these stable compounds provides the necessary energy to compensate for the high ionisation energies required to remove multiple electrons from the metal atom to achieve its highest oxidation state.

Other highly electronegative elements like chlorine are larger in size and less electronegative than fluorine and oxygen, making them less effective in stabilizing the highest oxidation states of metals.

Question 6

Q6. Which of the 3d series of transition metals exhibits the largest number of oxidation states and why?

Accepted Answer

Manganese (Mn) from the 3d series of transition metals exhibits the largest number of oxidation states.

**Explanation:**

1.  **Electronic Configuration:** Manganese has the electronic configuration [Ar] 3d⁵ 4s². 2.  **Availability of Electrons:** It has a total of seven valence electrons (two from 4s and five from 3d) that can be involved in bond formation. 3.  **Small Energy Difference:** The energy difference between the 3d and 4s orbitals is very small. This allows electrons from both the 4s and 3d orbitals to participate in bonding, leading to a wide range of oxidation states. 4.  **Oxidation States of Manganese:** Manganese exhibits oxidation states ranging from +2 to +7.
    *   **+2:** (e.g., MnSO₄, MnCl₂) - Involves loss of two 4s electrons, leaving a stable 3d⁵ configuration.
    *   **+3:** (e.g., Mn₂O₃)
    *   **+4:** (e.g., MnO₂, Pyrolusite)
    *   **+5:** (e.g., MnO₄³⁻)
    *   **+6:** (e.g., MnO₄²⁻, Manganate ion)
    *   **+7:** (e.g., Mn₂O₇, KMnO₄, Permanganate ion) - Involves the loss of all seven valence electrons (two 4s and five 3d electrons).

Other elements in the 3d series have fewer unpaired d-electrons or a larger energy gap between 3d and 4s orbitals, limiting their maximum oxidation states compared to manganese.

Question 7

Q7. What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Accepted Answer

Disproportionation is a specific type of redox reaction in which a single element in a particular oxidation state is simultaneously oxidized and reduced. This means that the same element acts as both the oxidizing agent and the reducing agent.

**Examples of disproportionation reactions in aqueous solution:**

1.  **Disproportionation of Copper(I) ion (Cu+):**
    In aqueous solution, Cu+ ion is unstable and undergoes disproportionation to form more stable Cu2+ and Cu metal.
    2Cu+(aq) → Cu2+(aq) + Cu(s)
    *   Here, copper in +1 oxidation state is oxidized to +2 oxidation state (in Cu2+) and simultaneously reduced to 0 oxidation state (in Cu metal).

2.  **Disproportionation of Manganate ion (MnO4^2-):**
    The manganate(VI) ion (MnO4^2-) is stable in alkaline medium but disproportionates in acidic or neutral solutions to form permanganate(VII) ion (MnO4-) and manganese dioxide (MnO2).
    3MnO4^2-(aq) + 4H+(aq) → 2MnO4-(aq) + MnO2(s) + 2H2O(l)
    *   Here, manganese in +6 oxidation state (in MnO4^2-) is oxidized to +7 oxidation state (in MnO4-) and simultaneously reduced to +4 oxidation state (in MnO2).

Question 8

Q8. Which metal in the first transition series exhibits +1 oxidation state most frequently and why?

Accepted Answer

The metal in the first transition series that exhibits the +1 oxidation state most frequently is **Copper (Cu)**.

**Reason:**
Copper has an anomalous electronic configuration of [Ar] 3d10 4s1. While most transition metals lose their 4s electrons first, copper can readily lose its single 4s electron to achieve a stable completely filled 3d10 configuration. This 3d10 configuration provides significant stability, making the Cu+ ion ([Ar] 3d10) a relatively common and stable oxidation state for copper, especially in compounds where the lattice energy or hydration energy is sufficient to stabilize it. Although Cu+ can disproportionate in aqueous solution (as discussed in Q7), it is a frequently observed oxidation state in many compounds.

Question 9

Q9. Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution?

Accepted Answer

Let's calculate the number of unpaired electrons for each gaseous ion first:

**1. Mn3+**
*   Atomic number of Mn = 25
*   Electronic configuration of Mn (ground state) = [Ar] 3d5 4s2
*   To form Mn3+, three electrons are removed (two from 4s and one from 3d).
*   Electronic configuration of Mn3+ = [Ar] 3d4
*   In 3d4, the electrons are distributed as: ↑ ↑ ↑ ↑ _
*   **Number of unpaired electrons = 4**

**2. Cr3+**
*   Atomic number of Cr = 24
*   Electronic configuration of Cr (ground state) = [Ar] 3d5 4s1 (anomalous)
*   To form Cr3+, three electrons are removed (one from 4s and two from 3d).
*   Electronic configuration of Cr3+ = [Ar] 3d3
*   In 3d3, the electrons are distributed as: ↑ ↑ ↑ _ _
*   **Number of unpaired electrons = 3**

**3. V3+**
*   Atomic number of V = 23
*   Electronic configuration of V (ground state) = [Ar] 3d3 4s2
*   To form V3+, three electrons are removed (two from 4s and one from 3d).
*   Electronic configuration of V3+ = [Ar] 3d2
*   In 3d2, the electrons are distributed as: ↑ ↑ _ _ _
*   **Number of unpaired electrons = 2**

**4. Ti3+**
*   Atomic number of Ti = 22
*   Electronic configuration of Ti (ground state) = [Ar] 3d2 4s2
*   To form Ti3+, three electrons are removed (two from 4s and one from 3d).
*   Electronic configuration of Ti3+ = [Ar] 3d1
*   In 3d1, the electron is distributed as: ↑ _ _ _ _
*   **Number of unpaired electrons = 1**

---

**Most stable in aqueous solution:**
Among Mn3+, Cr3+, V3+, and Ti3+, **Cr3+ is the most stable in aqueous solution.**

**Reason:**
In aqueous solution, transition metal ions are surrounded by water molecules (ligands), leading to crystal field splitting of the d-orbitals. For an octahedral complex (which is typical for hydrated ions), the five d-orbitals split into a lower energy t2g set (three orbitals) and a higher energy eg set (two orbitals).

*   **Cr3+ has a 3d3 configuration.** In an octahedral crystal field, these three electrons will occupy the three t2g orbitals (t2g3 eg0). This arrangement results in a half-filled t2g subshell, which is particularly stable due to maximum exchange energy and a significant Crystal Field Stabilization Energy (CFSE). The t2g3 configuration is considered very stable in octahedral complexes.

Let's compare the CFSE for low spin (as water is a weak field ligand, these will be high spin, but for d3, high spin and low spin are the same):
*   Ti3+ (d1): t2g1 eg0, CFSE = -0.4Δo
*   V3+ (d2): t2g2 eg0, CFSE = -0.8Δo
*   Cr3+ (d3): t2g3 eg0, CFSE = -1.2Δo (maximum CFSE for these ions)
*   Mn3+ (d4): t2g3 eg1, CFSE = -0.6Δo (for high spin)

The highest CFSE for Cr3+ (d3) contributes significantly to its thermodynamic stability in aqueous solution.

Question 10

Q10. Give examples and suggest reasons for the following features of the transition metal chemistry:
    (a) The lowest oxide of transition metal is basic, the highest is amphoteric/ac

Accepted Answer

Here are the explanations with examples for the given features of transition metal chemistry:

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

**Explanation:**
As the oxidation state of a transition metal increases, the covalent character of its oxide increases, and its basic character decreases, while its acidic character increases. This trend can be explained by Fajans' rules and the concept of electronegativity.

*   **Lower Oxidation States:** In lower oxidation states, the metal atom has a relatively low positive charge. This means it has a larger size and lower polarizing power. The metal-oxygen bond is predominantly ionic. Ionic oxides tend to be basic because the oxide ion (O2-) can readily accept protons (H+) from water to form hydroxide ions (OH-).
    *   **Example:** Manganese(II) oxide (MnO) is basic. It reacts with acids:
        MnO(s) + 2HCl(aq) → MnCl2(aq) + H2O(l)

*   **Higher Oxidation States:** In higher oxidation states, the metal atom has a high positive charge and a smaller effective size. This significantly increases its polarizing power, causing it to pull electron density from the oxygen atom more effectively. This leads to a greater covalent character in the metal-oxygen bond. Covalent oxides tend to be acidic because the highly electronegative oxygen atoms draw electron density away from the metal, making the metal-oxygen bond weaker and more susceptible to hydrolysis, releasing H+ ions.
    *   **Example:** Manganese(VII) oxide (Mn2O7) is highly acidic. It reacts with water to form permanganic acid:
        Mn2O7(l) + H2O(l) → 2HMnO4(aq)

*   **Intermediate Oxidation States:** Oxides in intermediate oxidation states often exhibit amphoteric behavior, meaning they can react with both acids and bases.
    *   **Example:** Chromium(III) oxide (Cr2O3) is amphoteric.
        Cr2O3(s) + 6HCl(aq) → 2CrCl3(aq) + 3H2O(l) (reacts as a base)
        Cr2O3(s) + 6NaOH(aq) + 3H2O(l) → 2Na3[Cr(OH)6](aq) (reacts as an acid)

### (b) A transition metal exhibits highest oxidation state in oxides and fluorides.

**Explanation:**
Oxygen and fluorine are the most electronegative elements in the periodic table and also have small atomic sizes. These properties allow them to stabilize the highest oxidation states of transition metals.

*   **High Electronegativity:** The high electronegativity of oxygen and fluorine means they have a strong tendency to attract electrons. When a transition metal forms bonds with these elements, the metal loses a large number of electrons (or shares them unequally), resulting in a high positive oxidation state.

*   **Small Size:** The small size of oxygen and fluorine atoms allows a large number of them to surround a central metal atom without significant steric hindrance. This enables the formation of multiple bonds (in the case of oxygen) or a large number of single bonds (in the case of fluorine), which is necessary to achieve and stabilize very high oxidation states.

*   **Example (Oxides):** Manganese exhibits its highest oxidation state of +7 in Mn2O7. Osmium exhibits +8 in OsO4.
    *   **Example (Fluorides):** Vanadium exhibits +5 in VF5. Chromium exhibits +6 in CrF6.

Other halogens (Cl, Br, I) are larger and less electronegative than F, so they cannot stabilize the highest oxidation states as effectively. For instance, the highest fluoride of manganese is MnF4, whereas its highest oxide is Mn2O7 (Mn in +7 state).

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

**Explanation:**
While oxides and fluorides can stabilize high oxidation states, oxoanions (anions containing oxygen and another element, like MnO4- or Cr2O7^2-) often allow metals to achieve their absolute highest oxidation states. This is primarily due to the ability of oxygen to form multiple bonds (pi bonds) with the metal, which helps to delocalize the electron density and stabilize the highly positive charge on the metal atom.

*   **Multiple Bonding (π-bonding):** Oxygen atoms can form both sigma (σ) and pi (π) bonds with the metal. The formation of multiple bonds (pπ-dπ bonding) allows the metal to achieve a very high formal oxidation state while effectively delocalizing the electron density, thus reducing the actual positive charge on the metal center. This strong covalent interaction helps to stabilize the high oxidation state.

*   **Charge Distribution:** In an oxoanion, the overall negative charge of the anion is distributed over the highly electronegative oxygen atoms, which further contributes to the stability of the central metal in a high oxidation state.

*   **Example:** Manganese exhibits its highest oxidation state of +7 in the permanganate ion (MnO4-).
    *   **Example:** Chromium exhibits its highest oxidation state of +6 in the chromate ion (CrO4^2-) and dichromate ion (Cr2O7^2-).

In these oxoanions, the metal atom is surrounded by several oxygen atoms, forming strong covalent bonds that stabilize the high positive charge on the metal.

Question 11

Q11. Complete the following equations:
    (i) Cr2O7^2- + 2OH- →
    (ii) MnO4^2- + 4H+ →
    (iii) Cr2O7^2- + 14H+ + 6Fe^2+ →
    (iv) MnO4- + 8H+ + 5Fe^2+ →
    (v) MnO4- + 8H+ +

Accepted Answer

Here are the completed and balanced chemical equations:

(i) **Cr2O7^2- + 2OH- → 2CrO4^2- + H2O**
    *   (Dichromate ion in basic medium converts to chromate ion)

(ii) **3MnO4^2- + 4H+ → 2MnO4- + MnO2 + 2H2O**
    *   (Manganate ion disproportionates in acidic medium)

(iii) **Cr2O7^2- + 14H+ + 6Fe^2+ → 2Cr3+ + 6Fe3+ + 7H2O**
    *   (Dichromate ion oxidizes Fe2+ to Fe3+ in acidic medium)

(iv) **MnO4- + 8H+ + 5Fe^2+ → Mn2+ + 5Fe3+ + 4H2O**
    *   (Permanganate ion oxidizes Fe2+ to Fe3+ in acidic medium)

(v) **2MnO4- + 16H+ + 5C2O4^2- → 2Mn2+ + 10CO2 + 8H2O**
    *   (Permanganate ion oxidizes oxalate ion to CO2 in acidic medium)

(vi) **2MnO4- + 16H+ + 10I- → 2Mn2+ + 5I2 + 8H2O**
    *   (Permanganate ion oxidizes iodide ion to iodine in acidic medium)

Question 12

Q12. What is the effect of pH on the dichromate ion solution?

Accepted Answer

The dichromate ion (Cr₂O₇²⁻) and chromate ion (CrO₄²⁻) exist in a pH-dependent equilibrium.

In acidic solution (low pH), the equilibrium shifts towards the formation of dichromate ions, and the solution appears orange:
2CrO₄²⁻ (yellow) + 2H⁺ ⇌ Cr₂O₇²⁻ (orange) + H₂O

In basic solution (high pH), the equilibrium shifts towards the formation of chromate ions, and the solution appears yellow:
Cr₂O₇²⁻ (orange) + 2OH⁻ ⇌ 2CrO₄²⁻ (yellow) + H₂O

Thus, the color of the solution changes from yellow (chromate) to orange (dichromate) upon acidification, and from orange (dichromate) to yellow (chromate) upon basification.

Question 13

Q13. How are the following conversions carried out:
(i) Chromite ore to potassium dichromate
(ii) Potassium dichromate to potassium chromate

Accepted Answer

### (i) Conversion of Chromite ore to Potassium Dichromate (K₂Cr₂O₇)

Potassium dichromate is prepared from chromite ore (FeCr₂O₄) in three main steps:

**Step 1: Preparation of Sodium Chromate from Chromite Ore**
Finely powdered chromite ore is mixed with sodium carbonate (or potassium carbonate) and lime (CaO) and roasted in a reverberatory furnace in the presence of excess air (oxygen). The iron oxidizes to ferric oxide, and chromium oxidizes to sodium chromate.

4FeCr₂O₄ (s) + 8Na₂CO₃ (s) + 7O₂ (g) → 8Na₂CrO₄ (s) + 2Fe₂O₃ (s) + 8CO₂ (g)

**Step 2: Conversion of Sodium Chromate to Sodium Dichromate**
The yellow solution of sodium chromate obtained in Step 1 is filtered and acidified with concentrated sulphuric acid. Sodium chromate, being basic, is converted to sodium dichromate (orange) in acidic medium.

2Na₂CrO₄ (aq) + H₂SO₄ (aq) → Na₂Cr₂O₇ (aq) + Na₂SO₄ (aq) + H₂O (l)

**Step 3: Conversion of Sodium Dichromate to Potassium Dichromate**
Sodium dichromate is more soluble than potassium dichromate. Therefore, a hot concentrated solution of sodium dichromate is treated with potassium chloride (KCl). On cooling, less soluble potassium dichromate crystallizes out.

Na₂Cr₂O₇ (aq) + 2KCl (aq) → K₂Cr₂O₇ (s) + 2NaCl (aq)

### (ii) Conversion of Potassium Dichromate to Potassium Chromate

Potassium dichromate (K₂Cr₂O₇) can be converted to potassium chromate (K₂CrO₄) by treating its solution with a strong base, such as potassium hydroxide (KOH) or sodium hydroxide (NaOH). The equilibrium between dichromate and chromate ions is pH-dependent. In alkaline medium, dichromate ions are converted to chromate ions.

K₂Cr₂O₇ (aq) + 2KOH (aq) → 2K₂CrO₄ (aq) + H₂O (l)

(Orange)                                (Yellow)

Question 14

Q14. Explain the following:
(i) Out of Cu⁺ and Cu²⁺, which ion is unstable in aqueous solution and why?
(ii) Why is Cr²⁺ reducing and Mn³⁺ oxidising when both have d⁴ configuration?

Accepted Answer

### (i) Out of Cu⁺ and Cu²⁺, which ion is unstable in aqueous solution and why?

**Cu⁺ ion is unstable in aqueous solution.**

This instability is due to a phenomenon called **disproportionation**. In an aqueous solution, Cu⁺ undergoes disproportionation, meaning it simultaneously oxidizes and reduces itself to form Cu²⁺ and Cu (metal), respectively.

**Reason:** The standard electrode potential for the Cu²⁺/Cu⁺ couple is +0.15 V, and for the Cu⁺/Cu couple is +0.52 V. The overall disproportionation reaction is:

2Cu⁺ (aq) → Cu²⁺ (aq) + Cu (s)

For this reaction, the standard electrode potential (E°) can be calculated as:
E° = E°(Cu²⁺/Cu⁺) - E°(Cu⁺/Cu) = (+0.15 V) - (+0.52 V) = -0.37 V

However, it's more appropriate to consider the standard Gibbs free energy change (ΔG°) for the disproportionation reaction.

For Cu⁺ → Cu²⁺ + e⁻, ΔG₁° = -nFE° = -(1)F(-0.15) = 0.15F
For Cu⁺ + e⁻ → Cu, ΔG₂° = -nFE° = -(1)F(0.52) = -0.52F

For 2Cu⁺ → Cu²⁺ + Cu, ΔG° = ΔG₁° + ΔG₂° = 0.15F - 0.52F = -0.37F

Since ΔG° is negative (-0.37F), the disproportionation reaction is spontaneous and thermodynamically favorable in aqueous solution. The higher hydration enthalpy of Cu²⁺ (due to its smaller size and higher charge) compared to Cu⁺ also plays a significant role in stabilizing the Cu²⁺ ion in aqueous solution, making the disproportionation of Cu⁺ favorable.

### (ii) Why is Cr²⁺ reducing and Mn³⁺ oxidising when both have d⁴ configuration?

Both Cr²⁺ and Mn³⁺ have a d⁴ electronic configuration.

*   **Cr²⁺ (d⁴) is a strong reducing agent:**
    Chromium (Cr) has an atomic number of 24, and its electronic configuration is [Ar] 3d⁵ 4s¹. Cr²⁺ has the configuration [Ar] 3d⁴. Cr²⁺ readily loses one electron to form Cr³⁺, which has a [Ar] 3d³ configuration. The 3d³ configuration in an octahedral ligand field is highly stable because it corresponds to a half-filled t₂g subshell (t₂g³ e_g⁰), which provides significant crystal field stabilization energy (CFSE). This strong tendency to achieve the stable Cr³⁺ state makes Cr²⁺ an excellent reducing agent.

*   **Mn³⁺ (d⁴) is an oxidising agent:**
    Manganese (Mn) has an atomic number of 25, and its electronic configuration is [Ar] 3d⁵ 4s². Mn³⁺ has the configuration [Ar] 3d⁴. Mn³⁺ readily gains one electron to form Mn²⁺, which has a [Ar] 3d⁵ configuration. The 3d⁵ configuration is highly stable because it corresponds to a half-filled d-orbital set. This strong tendency to achieve the stable Mn²⁺ state makes Mn³⁺ an excellent oxidizing agent.

In summary, the relative stability of the resulting oxidation states (Cr³⁺ with d³ and Mn²⁺ with d⁵) drives their respective reducing and oxidizing behaviors, even though both Cr²⁺ and Mn³⁺ start with a d⁴ configuration.

### (iii) Actinoids show more number of oxidation states than lanthanoids.

Actinoids (elements from Th to Lr) generally exhibit a wider range of oxidation states compared to lanthanoids (elements from Ce to Lu).

**Reasons:**

1.  **Comparable Energies of Orbitals:** In actinoids, the 5f, 6d, and 7s orbitals are of comparable energies. This means that electrons from all these orbitals can participate in bonding, leading to a greater variety of oxidation states. For example, uranium (U) shows +3, +4, +5, and +6 oxidation states, and plutonium (Pu) shows +3, +4, +5, +6, and +7.
2.  **Poor Shielding by 5f Electrons:** The 5f electrons in actinoids are less effectively shielded from the nuclear charge compared to the 4f electrons in lanthanoids. This results in the 5f electrons being more exposed and available for bonding. While 4f electrons are deeply buried and primarily involved in shielding, 5f electrons are more accessible.

In contrast, for lanthanoids, the 4f orbitals are much deeper inside the atom and are well-shielded by the 5s and 5p orbitals. Consequently, only the 5d and 6s electrons primarily participate in bonding, leading to a predominant +3 oxidation state, with +2 and +4 being less common and less stable.

### (iv) Transition metals and their compounds are generally found to be good catalysts.

Transition metals and their compounds exhibit excellent catalytic properties due to the following reasons:

1.  **Variable Oxidation States:** Transition metals can exhibit multiple oxidation states. This allows them to form unstable intermediate compounds with reactants, providing an alternative reaction pathway with a lower activation energy. The catalyst can easily switch between oxidation states during the reaction cycle.
2.  **Large Surface Area:** Many transition metals and their compounds (especially in finely divided states) have large surface areas. This provides numerous active sites for adsorption of reactant molecules, bringing them into close proximity and facilitating bond formation or breaking.
3.  **Ability to Form Intermediates:** Transition metals can form temporary bonds with reactant molecules (adsorption) through their vacant d-orbitals. This weakens the bonds within the reactant molecules and facilitates their reaction. After the reaction, the products desorb, and the catalyst is regenerated.
4.  **Formation of Interstitial Compounds:** The ability to form interstitial compounds (where small atoms occupy voids in the metal lattice) can also contribute to catalytic activity by modifying the electronic and structural properties of the metal surface.

**Examples:**
*   Vanadium(V) oxide (V₂O₅) in the contact process for H₂SO₄ production.
*   Finely divided iron (Fe) in the Haber process for NH₃ synthesis.
*   Nickel (Ni) in hydrogenation reactions.

### (v) Interstitial compounds are well known for transition metals.

**Interstitial compounds** are non-stoichiometric compounds formed when small atoms like hydrogen (H), carbon (C), nitrogen (N), or boron (B) get trapped in the interstitial voids (spaces) within the crystal lattice of transition metals.

**Reasons for their formation by transition metals:**

1.  **Appropriate Atomic Size:** Transition metals have relatively large atomic sizes and an open crystal lattice structure (e.g., face-centered cubic, body-centered cubic, hexagonal close-packed). This creates suitable voids or interstitial sites that can accommodate small non-metal atoms.
2.  **Metallic Bonding:** The strong metallic bonding in transition metals is not significantly disrupted by the inclusion of these small atoms. Instead, the non-metal atoms form covalent bonds with the metal atoms, further strengthening the lattice.

**Characteristics of Interstitial Compounds:**

*   **High Melting Points:** Higher than the pure metals due to stronger bonding.
*   **Hardness:** Very hard, some are as hard as diamond (e.g., tungsten carbides).
*   **Chemical Inertness:** Chemically unreactive.
*   **Retain Metallic Conductivity:** They generally retain their metallic conductivity.
*   **Non-stoichiometric:** Their composition is often non-stoichiometric (e.g., TiH₁.₇, Fe₃C, Mn₄N).

Question 15

Q15. Describe the preparation of potassium permanganate from pyrolusite ore. Write the chemical reactions involved.

Accepted Answer

Potassium permanganate (KMnO₄) is prepared from pyrolusite ore (MnO₂) in two main steps:

### Step 1: Conversion of Pyrolusite ore (MnO₂) to Potassium Manganate (K₂MnO₄)

Finely powdered pyrolusite ore (MnO₂) is fused with an alkali (like KOH) in the presence of an oxidizing agent (like air, KNO₃, or KClO₃). This process oxidizes manganese from its +4 oxidation state in MnO₂ to the +6 oxidation state in potassium manganate (K₂MnO₄). The fusion is typically carried out at high temperatures.

**Reaction:**
2MnO₂ (s) + 4KOH (s) + O₂ (g) → 2K₂MnO₄ (s) + 2H₂O (l)
(Pyrolusite)                                (Potassium Manganate, Green)

Alternatively, if an oxidizing agent like KNO₃ is used:
MnO₂ (s) + 2KOH (s) + KNO₃ (s) → K₂MnO₄ (s) + KNO₂ (s) + H₂O (l)

The fused mass is then extracted with water to obtain a green solution of potassium manganate.

### Step 2: Oxidation of Potassium Manganate (K₂MnO₄) to Potassium Permanganate (KMnO₄)

Potassium manganate (K₂MnO₄), where manganese is in the +6 oxidation state, is then oxidized to potassium permanganate (KMnO₄), where manganese is in the +7 oxidation state. This can be achieved by chemical oxidation, electrolytic oxidation, or disproportionation in acidic medium.

**Method A: Electrolytic Oxidation**
This is the most common industrial method. The green solution of potassium manganate is electrolyzed. Oxidation occurs at the anode, converting manganate ions to permanganate ions.

At anode: 2K₂MnO₄ (aq) + 2H₂O (l) → 2KMnO₄ (aq) + 2KOH (aq) + H₂ (g)
(Green)                                    (Purple)

Ionic reaction at anode: 2MnO₄²⁻ (aq) → 2MnO₄⁻ (aq) + 2e⁻

**Method B: Chemical Oxidation**
Potassium manganate can be oxidized by strong oxidizing agents like chlorine (Cl₂) or ozone (O₃).

2K₂MnO₄ (aq) + Cl₂ (g) → 2KMnO₄ (aq) + 2KCl (aq)

**Method C: Disproportionation in Acidic Medium**
Potassium manganate is stable in alkaline and neutral solutions, but in acidic solutions, it disproportionates into permanganate and manganese dioxide.

3K₂MnO₄ (aq) + 4H⁺ (aq) → 2KMnO₄ (aq) + MnO₂ (s) + 2H₂O (l) + 2K⁺ (aq)

Ionic reaction: 3MnO₄²⁻ (aq) + 4H⁺ (aq) → 2MnO₄⁻ (aq) + MnO₂ (s) + 2H₂O (l)

After oxidation, the purple solution of potassium permanganate is concentrated by evaporation and then cooled to obtain crystals of KMnO₄.

Question 16

Q16. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Accepted Answer

### What are Alloys?
An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal, which has metallic properties. Alloys are typically formed by melting the constituent elements together and then cooling the mixture. They are designed to have properties that are superior to those of their individual components, such as increased hardness, strength, corrosion resistance, or specific electrical/magnetic properties.

### Important Alloy Containing Lanthanoid Metals: Mischmetal
An important alloy that contains some of the lanthanoid metals is **Mischmetal**.

**Composition:** Mischmetal typically consists of about 95% lanthanoid metals (mainly cerium (Ce) ~50%, lanthanum (La) ~25%, neodymium (Nd) ~18%, praseodymium (Pr) ~5%) and about 5% iron (Fe), along with traces of sulfur, carbon, calcium, and aluminium.

**Uses of Mischmetal:**
1.  **Lighter Flints:** Its most common use is in the production of flints for lighters and toys, where it produces sparks when struck.
2.  **Metallurgical Applications:** It is used in metallurgy to deoxidize molten metals and to remove impurities. For example, it is added to magnesium-based alloys to improve their strength and workability.
3.  **Bullet and Shell Production:** It is used in the production of tracer bullets and shells, where the burning of lanthanoid metals produces a bright flame.

Question 17

Q17. Compare the chemistry of the actinoids with that of lanthanoids with special reference to:
(a) electronic configuration
(b) oxidation states
(c) chemical reactivity and
(d) mag

Accepted Answer

The chemistry of actinoids and lanthanoids, while sharing some similarities as f-block elements, also exhibits significant differences due to the nature of their f-orbitals and relativistic effects. Here's a comparison:

### (a) Electronic Configuration

**Lanthanoids:**
*   General electronic configuration: [Xe] 4f^(1-14) 5d^(0-1) 6s^2.
*   The filling of 4f orbitals occurs after 6s and sometimes 5d. The 4f orbitals are deeply buried within the atom and are well shielded by the 5s and 5p orbitals. This shielding makes them less available for bonding.
*   Examples: La (Z=57): [Xe] 5d^1 6s^2; Ce (Z=58): [Xe] 4f^1 5d^1 6s^2; Gd (Z=64): [Xe] 4f^7 5d^1 6s^2; Lu (Z=71): [Xe] 4f^14 5d^1 6s^2.

**Actinoids:**
*   General electronic configuration: [Rn] 5f^(0-14) 6d^(0-2) 7s^2.
*   The filling of 5f orbitals occurs after 7s and sometimes 6d. The 5f orbitals are less shielded than 4f orbitals and extend further out into space, making them more accessible for bonding.
*   Examples: Ac (Z=89): [Rn] 6d^1 7s^2; Th (Z=90): [Rn] 6d^2 7s^2; Pa (Z=91): [Rn] 5f^2 6d^1 7s^2; U (Z=92): [Rn] 5f^3 6d^1 7s^2.

### (b) Oxidation States

**Lanthanoids:**
*   The most common and stable oxidation state is +3. This is due to the loss of two 6s electrons and one 5d or 4f electron.
*   Some elements also exhibit +2 and +4 oxidation states, particularly when they lead to stable f^0, f^7, or f^14 configurations. For example, Eu^(2+) (4f^7), Yb^(2+) (4f^14), Ce^(4+) (4f^0), Tb^(4+) (4f^7).
*   The +2 and +4 states are generally less stable than +3 and often act as reducing or oxidizing agents, respectively.

**Actinoids:**
*   They exhibit a wider range of oxidation states compared to lanthanoids, primarily due to the comparable energies of 5f, 6d, and 7s orbitals, and the less effective shielding of 5f electrons.
*   The common oxidation states are +3, +4, +5, +6, and +7. For example, Uranium shows +3, +4, +5, +6; Neptunium shows +3, +4, +5, +6, +7; Plutonium shows +3, +4, +5, +6, +7.
*   The +3 oxidation state becomes more stable as we move across the actinoid series (actinoid contraction), similar to lanthanoids.

### (c) Chemical Reactivity

**Lanthanoids:**
*   All lanthanoids are highly reactive metals, comparable to alkaline earth metals.
*   They readily tarnish in air, react with water to form hydroxides and hydrogen, react with acids to liberate hydrogen, and react with halogens, sulfur, nitrogen, and carbon at elevated temperatures.
*   Their reactivity generally decreases slightly across the series due to the lanthanoid contraction.

**Actinoids:**
*   Actinoids are even more reactive than lanthanoids, especially in their early part of the series. This is due to the greater availability of 5f electrons for bonding.
*   They are highly electropositive and react vigorously with most non-metals. They are very reactive towards oxygen and water.
*   Many actinoids are pyrophoric (ignite spontaneously in air) in finely divided form.
*   They form compounds with various non-metals like halogens, sulfur, nitrogen, and carbon.
*   Most actinoids are radioactive, which also influences their handling and study.

### (d) Magnetic Properties

**Lanthanoids:**
*   Most lanthanoid ions (except those with f^0 and f^14 configurations, i.e., La^(3+), Ce^(4+), Yb^(2+), Lu^(3+)) are paramagnetic due to the presence of unpaired electrons in the 4f orbitals.
*   The 4f electrons are well-shielded, so their orbital motion contributes significantly to the total magnetic moment. The magnetic moments are calculated using the formula μ = g√[J(J+1)], where J is the total angular momentum quantum number.
*   The magnetic moments are generally higher than those of transition metals.

**Actinoids:**
*   Actinoid ions are also generally paramagnetic due to unpaired 5f electrons.
*   However, the 5f electrons are less shielded and are more involved in bonding. This leads to quenching of the orbital contribution to the magnetic moment, especially for the earlier actinoids.
*   The magnetic moments are often less than the calculated values based on spin-only or even spin-orbital coupling models, and they are generally lower than those of corresponding lanthanoids.
*   The magnetic properties are more complex and difficult to interpret due to the greater delocalization of 5f electrons and stronger spin-orbit coupling.

**Summary Table:**

Question 18

Q18. Discuss the position of lanthanoids in the periodic table.

Accepted Answer

The lanthanoids (also known as lanthanides) are a series of 14 elements from Cerium (Ce, Z=58) to Lutetium (Lu, Z=71). They are formally placed in **Group 3** and **Period 6** of the periodic table, immediately after Lanthanum (La, Z=57).

**Reason for their separate placement:**

1.  **Electronic Configuration:** In lanthanoids, the differentiating electron enters the antepenultimate 4f subshell. If these 14 elements were placed within the main body of the periodic table, it would disrupt the regular progression of properties of the d-block elements and make the table excessively long and cumbersome.
2.  **Similar Properties:** All lanthanoids exhibit very similar chemical and physical properties, primarily due to the very small change in atomic and ionic radii across the series (lanthanoid contraction) and their predominant +3 oxidation state. Placing them together in a separate block highlights this chemical similarity.
3.  **To Maintain Order:** To maintain the logical structure and periodicity of the main body of the periodic table, especially the d-block and p-block elements, the lanthanoids are conventionally placed as a separate row at the bottom of the periodic table. This arrangement allows the periodic table to remain compact and visually clear, while still acknowledging their position after Lanthanum in the sixth period.

Therefore, while they belong to Group 3, Period 6, they are typically shown as a separate row below the main table for convenience and to emphasize their unique f-block characteristics.

Question 19

Q19. What are the different oxidation states exhibited by the lanthanoids?

Accepted Answer

The most common and stable oxidation state exhibited by all lanthanoids is **+3**.

In addition to the +3 oxidation state, some lanthanoids also show **+2** and **+4** oxidation states. These less common oxidation states are generally observed when they lead to a more stable electronic configuration, specifically:

*   **f^0 configuration:** e.g., Ce^(4+) (4f^0)
*   **f^7 configuration:** e.g., Eu^(2+) (4f^7), Tb^(4+) (4f^7)
*   **f^14 configuration:** e.g., Yb^(2+) (4f^14)

**Examples of Lanthanoids and their Oxidation States:**

*   **Cerium (Ce):** +3, +4 (Ce^(4+) is a strong oxidizing agent, stable in solution)
*   **Praseodymium (Pr):** +3, +4
*   **Neodymium (Nd):** +3, +2
*   **Samarium (Sm):** +3, +2 (Sm^(2+) is a strong reducing agent)
*   **Europium (Eu):** +3, +2 (Eu^(2+) is stable due to 4f^7 configuration, strong reducing agent)
*   **Terbium (Tb):** +3, +4 (Tb^(4+) is stable due to 4f^7 configuration)
*   **Dysprosium (Dy):** +3, +4
*   **Ytterbium (Yb):** +3, +2 (Yb^(2+) is stable due to 4f^14 configuration, strong reducing agent)

While +2 and +4 states exist, they are often less stable than +3 and tend to revert to the +3 state by acting as reducing or oxidizing agents, respectively. For instance, Ce^(4+) is a strong oxidizing agent, and Eu^(2+) and Yb^(2+) are strong reducing agents.

Question 20

Q20. Explain the following:
(a) Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why?
(b) Actonoid contraction is greater from element to element than lanthanoid con

Accepted Answer

### (a) Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why?

**Cu+ (cuprous ion) is unstable in aqueous solution and undergoes disproportionation.**

**Explanation:**

Copper (Cu) can exist in two common oxidation states: +1 (Cu+) and +2 (Cu2+). In an aqueous solution, Cu+ ions are unstable and tend to disproportionate into Cu2+ ions and metallic copper (Cu). This means that one Cu+ ion is oxidized to Cu2+, and another Cu+ ion is reduced to Cu.

The disproportionation reaction is:
2Cu+(aq) → Cu2+(aq) + Cu(s)

**Reason for instability:**

The instability of Cu+ in aqueous solution is primarily due to the much higher hydration enthalpy of Cu2+ compared to Cu+. Although the ionization enthalpy required to remove the second electron from Cu+ to form Cu2+ is high, this energy is more than compensated by the significantly greater hydration enthalpy of the Cu2+ ion.

*   **Hydration Enthalpy:** Cu2+ has a smaller size and a higher charge (+2) compared to Cu+ (larger size, +1 charge). According to Fajan's rules and general principles of ionic interactions, smaller ions with higher charges have a greater charge density and thus interact more strongly with water molecules. This leads to a much larger release of energy upon hydration (more negative hydration enthalpy) for Cu2+ than for Cu+.
*   **Lattice Energy (for solid Cu+ compounds):** While solid Cu+ compounds (like CuCl) are stable due to high lattice energy, when dissolved in water, the hydration energy factor becomes dominant.

The overall energy change for the disproportionation reaction (2Cu+(aq) → Cu2+(aq) + Cu(s)) is favorable (negative ΔG), making it a spontaneous process in aqueous solution. The high hydration energy of Cu2+ drives this reaction.

### (b) Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

**Actinoid contraction is greater from element to element than lanthanoid contraction.**

**Explanation:**

Both lanthanoids and actinoids exhibit a gradual decrease in atomic and ionic radii across their respective series, known as lanthanoid contraction and actinoid contraction. This contraction is due to the poor shielding effect of the f-electrons. However, actinoid contraction is significantly more pronounced than lanthanoid contraction for the following reasons:

1.  **Poorer Shielding of 5f Electrons:** The 5f orbitals in actinoids are more diffuse and extend further out into space compared to the 4f orbitals in lanthanoids. This means that the 5f electrons provide even poorer shielding of the nuclear charge from the outermost electrons than the 4f electrons. As a result, the effective nuclear charge experienced by the outermost electrons increases more significantly from element to element in actinoids, leading to a stronger pull towards the nucleus and a greater reduction in size.
2.  **Greater Number of Unpaired Electrons and Relativistic Effects:** While not the primary reason, the greater number of unpaired electrons in 5f orbitals (especially in the early actinoids) and the relativistic effects (which become more significant for heavier elements with high nuclear charge, affecting the energies and sizes of orbitals) also contribute to the more pronounced contraction in actinoids.
3.  **Higher Principal Quantum Number:** The 5f orbitals are in a higher principal quantum shell (n=5) compared to 4f orbitals (n=4). Electrons in higher principal quantum shells are generally less effectively shielded by inner electrons, contributing to a stronger effective nuclear charge increase across the series.

In essence, the 5f electrons are less effective at shielding the increasing nuclear charge than the 4f electrons, leading to a stronger and more cumulative increase in effective nuclear charge across the actinoid series. This stronger effective nuclear charge results in a greater inward pull on the valence electrons, causing a more significant decrease in atomic and ionic radii from one actinoid element to the next.

Question 21

Q21. What are the general characteristics of the transition elements and in which of the following aspects do the transition elements differ from the non-transition elements:
(a) elect

Accepted Answer

### General Characteristics of Transition Elements

Transition elements are d-block elements located between the s-block and p-block in the periodic table (Groups 3-12). They are characterized by incompletely filled d-orbitals in their atomic state or in any of their common oxidation states. Their general characteristics include:

1.  **Metallic Character:** All transition elements are typical metals. They are hard, have high tensile strength, ductility, malleability, high thermal and electrical conductivity, and metallic luster.
2.  **High Melting and Boiling Points:** Due to strong metallic bonding (involving both s and d electrons), they have high melting and boiling points.
3.  **Variable Oxidation States:** They exhibit a wide range of oxidation states due to the participation of both (n-1)d and ns electrons in bonding.
4.  **Formation of Coloured Ions:** Most transition metal ions are coloured in solid or solution states due to d-d transitions (absorption of visible light and emission of complementary colour).
5.  **Paramagnetism:** Many transition metal ions are paramagnetic due to the presence of unpaired electrons in their (n-1)d orbitals.
6.  **Catalytic Properties:** Many transition metals and their compounds act as good catalysts (e.g., Fe in Haber process, Ni in hydrogenation) due to their variable oxidation states and ability to form intermediate compounds, and large surface area.
7.  **Formation of Interstitial Compounds:** They form interstitial compounds with small non-metal atoms (H, C, N, B) trapped in the interstitial sites of their crystal lattices. These compounds are hard, have high melting points, and retain metallic conductivity.
8.  **Formation of Alloys:** They readily form alloys with other transition metals and non-transition metals due to similar atomic sizes.
9.  **Complex Formation:** They have a strong tendency to form coordination compounds (complexes) due to their small size, high ionic charge, and availability of vacant d-orbitals to accept lone pairs of electrons from ligands.

### Differences from Non-Transition Elements

Transition elements differ significantly from non-transition elements (s-block and p-block elements) in several key aspects:

#### (a) Electronic Configuration

*   **Transition Elements:** Have a general electronic configuration of (n-1)d^(1-10) ns^(1-2). They are characterized by incompletely filled (n-1)d orbitals in their ground state or in one of their common oxidation states. The differentiating electron enters the (n-1)d subshell.
*   **Non-Transition Elements:**
    *   **s-block elements:** Have a general electronic configuration of ns^(1-2). Their outermost shell has 1 or 2 electrons, and there are no d-electrons involved in bonding.
    *   **p-block elements:** Have a general electronic configuration of ns^2 np^(1-6). Their outermost shell contains 3 to 8 electrons, and the differentiating electron enters the np subshell. They do not have partially filled d-orbitals in their valence shell.

#### (b) Enthalpy of Atomisation

*   **Transition Elements:** Have very high enthalpies of atomisation. This is due to the strong metallic bonding resulting from the involvement of a large number of unpaired electrons in the (n-1)d orbitals, in addition to the ns electrons, in forming metallic bonds. The more unpaired d-electrons, the stronger the metallic bond and thus higher the enthalpy of atomisation.
*   **Non-Transition Elements:**
    *   **s-block elements:** Have relatively low enthalpies of atomisation. They typically have only 1 or 2 valence electrons (ns electrons) participating in metallic bonding, which results in weaker metallic bonds compared to transition metals.
    *   **p-block elements:** Their enthalpies of atomisation vary widely. Metals in the p-block (e.g., Al, Ga, Sn, Pb) have moderate enthalpies of atomisation, generally lower than transition metals, as their bonding primarily involves p and s electrons. Non-metals in the p-block (e.g., N2, O2, F2, Cl2) exist as discrete molecules with weak intermolecular forces, leading to very low enthalpies of atomisation (or sublimation/vaporization).

#### (c) Oxidation States

*   **Transition Elements:** Exhibit a wide range of variable oxidation states. This is because the (n-1)d and ns orbitals are very close in energy, allowing both sets of electrons to participate in bonding. For example, Manganese (Mn) shows oxidation states from +2 to +7.
*   **Non-Transition Elements:**
    *   **s-block elements:** Typically show only one characteristic oxidation state (e.g., Group 1 elements show +1, Group 2 elements show +2). This is because they lose all their valence s-electrons to achieve a stable noble gas configuration.
    *   **p-block elements:** Show a limited number of oxidation states, often differing by two units (due to inert pair effect for heavier elements) or by one unit (e.g., Group 13: +3; Group 14: +2, +4; Group 15: -3, +3, +5). Their oxidation states are generally determined by the number of valence electrons in the ns and np subshells, and the energy difference between these orbitals is usually larger than that between (n-1)d and ns orbitals in transition metals.

Question 22

Q22. What is the effect of increasing pH on the dichromate ion?

Accepted Answer

The dichromate ion (Cr2O7^2-) exists in equilibrium with the chromate ion (CrO4^2-). This equilibrium is pH-dependent.

In acidic medium (low pH), dichromate ions are stable:
2CrO4^2- (yellow) + 2H^+  ⇌ Cr2O7^2- (orange) + H2O

When the pH is increased (i.e., the solution becomes more alkaline or basic), the H^+ ion concentration decreases. According to Le Chatelier's principle, the equilibrium will shift to the left to consume the added OH^- (or to produce H^+ by shifting left), favoring the formation of chromate ions.

Cr2O7^2- (orange) + 2OH^-  ⇌ 2CrO4^2- (yellow) + H2O

Therefore, increasing the pH (making the solution alkaline) converts the orange dichromate ion into the yellow chromate ion.

Question 23

Q23. How are the following conversions carried out:
(i) Chromite ore to potassium dichromate
(ii) Potassium dichromate to chromium sulphate

Accepted Answer

### (i) Conversion of Chromite ore to Potassium Dichromate (K2Cr2O7)

The conversion of chromite ore (FeCr2O4) to potassium dichromate involves three main steps:

**Step 1: Preparation of Sodium Chromate from Chromite Ore**
Finely powdered chromite ore is mixed with sodium carbonate (Na2CO3) and quicklime (CaO) and roasted in a reverberatory furnace in the presence of excess air (oxygen) at a high temperature (around 1000-1200 °C). The iron present in the ore is oxidized to ferric oxide, and chromium is oxidized to sodium chromate.

4FeCr2O4 (s) + 8Na2CO3 (s) + 7O2 (g) → 8Na2CrO4 (s) + 2Fe2O3 (s) + 8CO2 (g)

The roasted mass is then extracted with water. Sodium chromate (Na2CrO4) is soluble in water, while ferric oxide (Fe2O3) is insoluble and can be filtered off.

**Step 2: Conversion of Sodium Chromate to Sodium Dichromate**
The yellow solution of sodium chromate is treated with concentrated sulfuric acid (H2SO4). Sodium chromate is converted into orange sodium dichromate (Na2Cr2O7) in the acidic medium.

2Na2CrO4 (aq) + H2SO4 (aq) → Na2Cr2O7 (aq) + Na2SO4 (aq) + H2O (l)

Sodium dichromate is more soluble than sodium sulfate. On cooling, sodium sulfate crystallizes out and is removed by filtration. The concentrated solution of sodium dichromate is then obtained.

**Step 3: Conversion of Sodium Dichromate to Potassium Dichromate**
Sodium dichromate is then treated with potassium chloride (KCl). Due to the lower solubility of potassium dichromate (K2Cr2O7) compared to sodium dichromate, K2Cr2O7 crystallizes out on cooling.

Na2Cr2O7 (aq) + 2KCl (aq) → K2Cr2O7 (s) + 2NaCl (aq)

Potassium dichromate crystals are separated by filtration and purified by recrystallization.

### (ii) Conversion of Potassium Dichromate to Chromium Sulphate (Cr2(SO4)3)

Potassium dichromate (K2Cr2O7) acts as a strong oxidizing agent in acidic medium. When it reacts with a reducing agent in the presence of sulfuric acid, the chromium (VI) in dichromate is reduced to chromium (III) ions, which then form chromium sulfate.

One common method involves reducing potassium dichromate with a suitable reducing agent like sulfur dioxide (SO2) or ethanol (C2H5OH) in the presence of dilute sulfuric acid.

**Using Sulfur Dioxide (SO2) as a Reducing Agent:**
In acidic medium, dichromate ions are reduced to Cr^3+ ions. Sulfur dioxide is oxidized to sulfate ions (SO4^2-).

K2Cr2O7 (aq) + 3SO2 (g) + H2SO4 (aq) → K2SO4 (aq) + Cr2(SO4)3 (aq) + H2O (l)

Alternatively, in ionic form:
Cr2O7^2- (aq) + 3SO2 (g) + 2H^+ (aq) → 2Cr^3+ (aq) + 3SO4^2- (aq) + H2O (l)

The resulting solution contains chromium(III) sulfate, which can be isolated by evaporation and crystallization.

Question 24

Q24. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(a) iodide
(b) iron(II) solution
(c) H2S

Accepted Answer

### Oxidising Action of Potassium Dichromate (K2Cr2O7)

Potassium dichromate (K2Cr2O7) is a powerful oxidizing agent in acidic medium. In an acidic solution, the dichromate ion (Cr2O7^2-) accepts electrons and gets reduced from chromium in the +6 oxidation state to chromium in the +3 oxidation state (Cr^3+ ions). The standard electrode potential for this reduction is +1.33 V, indicating its strong oxidizing power.

The half-reaction for the reduction of dichromate ion in acidic medium is:
Cr2O7^2- (aq) + 14H^+ (aq) + 6e^- → 2Cr^3+ (aq) + 7H2O (l)

This reduction causes the orange color of the dichromate solution to change to the green color of Cr^3+ ions. The reducing agent itself gets oxidized in the process.

### Ionic Equations for Reactions:

**(a) Reaction with Iodide (I^-) ions**

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

*   **Oxidation half-reaction:** 2I^- (aq) → I2 (s) + 2e^-
*   **Reduction half-reaction:** Cr2O7^2- (aq) + 14H^+ (aq) + 6e^- → 2Cr^3+ (aq) + 7H2O (l)

To balance the electrons, multiply the oxidation half-reaction by 3:
3(2I^- (aq) → I2 (s) + 2e^-)  =>  6I^- (aq) → 3I2 (s) + 6e^-

**Overall Ionic Equation:**
Cr2O7^2- (aq) + 6I^- (aq) + 14H^+ (aq) → 2Cr^3+ (aq) + 3I2 (s) + 7H2O (l)

**(b) Reaction with Iron(II) solution (Fe^2+ ions)**

Potassium dichromate oxidizes iron(II) ions (Fe^2+) to iron(III) ions (Fe^3+).

*   **Oxidation half-reaction:** Fe^2+ (aq) → Fe^3+ (aq) + e^-
*   **Reduction half-reaction:** Cr2O7^2- (aq) + 14H^+ (aq) + 6e^- → 2Cr^3+ (aq) + 7H2O (l)

To balance the electrons, multiply the oxidation half-reaction by 6:
6(Fe^2+ (aq) → Fe^3+ (aq) + e^-)  =>  6Fe^2+ (aq) → 6Fe^3+ (aq) + 6e^-

**Overall Ionic Equation:**
Cr2O7^2- (aq) + 6Fe^2+ (aq) + 14H^+ (aq) → 2Cr^3+ (aq) + 6Fe^3+ (aq) + 7H2O (l)

**(c) Reaction with H2S (Hydrogen Sulfide)**

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

*   **Oxidation half-reaction:** H2S (g) → S (s) + 2H^+ (aq) + 2e^-
*   **Reduction half-reaction:** Cr2O7^2- (aq) + 14H^+ (aq) + 6e^- → 2Cr^3+ (aq) + 7H2O (l)

To balance the electrons, multiply the oxidation half-reaction by 3:
3(H2S (g) → S (s) + 2H^+ (aq) + 2e^-)  =>  3H2S (g) → 3S (s) + 6H^+ (aq) + 6e^-

**Overall Ionic Equation:**
Cr2O7^2- (aq) + 3H2S (g) + 14H^+ (aq) → 2Cr^3+ (aq) + 3S (s) + 6H^+ (aq) + 7H2O (l)

Simplifying the H^+ ions (14H^+ on LHS and 6H^+ on RHS):
Cr2O7^2- (aq) + 3H2S (g) + 8H^+ (aq) → 2Cr^3+ (aq) + 3S (s) + 7H2O (l)

Question 25

Q25. Describe the preparation of potassium permanganate from pyrolusite ore. Write the chemical reactions involved.

Accepted Answer

### Preparation of Potassium Permanganate (KMnO4) from Pyrolusite Ore (MnO2)

Potassium permanganate is prepared from pyrolusite ore (MnO2) in two main steps:

**Step 1: Conversion of Pyrolusite Ore to Potassium Manganate (K2MnO4)**

Finely powdered pyrolusite ore (MnO2) is fused with an alkali metal hydroxide (like KOH) and an oxidizing agent (like KNO3 or air/O2) at high temperatures (around 800-900 °C). This process oxidizes manganese(IV) in MnO2 to manganese(VI) in potassium manganate (K2MnO4), which is green in color.

2MnO2 (s) + 4KOH (s) + O2 (g) → 2K2MnO4 (s) + 2H2O (l)

Alternatively, if an oxidizing agent like KNO3 is used:
3MnO2 (s) + 6KOH (s) + KNO3 (s) → 3K2MnO4 (s) + KNO2 (s) + 3H2O (l)

The fused green mass containing potassium manganate is then extracted with water.

**Step 2: Oxidation of Potassium Manganate to Potassium Permanganate (KMnO4)**

The green solution of potassium manganate (K2MnO4) is then oxidized to purple potassium permanganate (KMnO4). This can be achieved by chemical oxidation or electrolytic oxidation.

**Method A: Chemical Oxidation**
Potassium manganate (MnO4^2-, Mn in +6 oxidation state) can be oxidized to potassium permanganate (MnO4^-, Mn in +7 oxidation state) using various oxidizing agents like chlorine, ozone, or even by disproportionation in an acidic or neutral medium.

*   **Using Chlorine (Cl2):**
    2K2MnO4 (aq) + Cl2 (g) → 2KMnO4 (aq) + 2KCl (aq)
    (Ionic: 2MnO4^2- (aq) + Cl2 (g) → 2MnO4^- (aq) + 2Cl^- (aq))

*   **Using Ozone (O3):**
    2K2MnO4 (aq) + O3 (g) + H2O (l) → 2KMnO4 (aq) + 2KOH (aq) + O2 (g)
    (Ionic: 2MnO4^2- (aq) + O3 (g) + H2O (l) → 2MnO4^- (aq) + 2OH^- (aq) + O2 (g))

*   **Disproportionation in Acidic/Neutral Medium:**
    This is the most common industrial method. When the green manganate solution is acidified (e.g., with dilute H2SO4 or CO2), the manganate ion disproportionates into permanganate and manganese dioxide.
    3MnO4^2- (aq) + 4H^+ (aq) → 2MnO4^- (aq) + MnO2 (s) + 2H2O (l)
    (The MnO2 formed can be recycled)

**Method B: Electrolytic Oxidation**
This is the preferred industrial method. The green solution of potassium manganate is electrolyzed. Manganate ions are oxidized to permanganate ions at the anode.

At Anode: MnO4^2- (aq) → MnO4^- (aq) + e^-
At Cathode: 2H2O (l) + 2e^- → H2 (g) + 2OH^- (aq)

Potassium permanganate crystallizes out from the solution as dark purple, almost black, crystals. It is then filtered, washed, and dried.

Question 26

Q26. What are the similarities and differences between the chemistry of lanthanoids and actinoids?

Accepted Answer

### Similarities and Differences between Lanthanoids and Actinoids

Lanthanoids (4f-series) and Actinoids (5f-series) are two series of f-block elements. While they share some general characteristics due to their f-electron filling, they also exhibit significant differences.

#### Similarities:

1.  **Electronic Configuration:** Both series involve the filling of f-orbitals (4f for lanthanoids and 5f for actinoids). Their general electronic configuration is (n-2)f^(1-14) (n-1)d^(0-1) ns^2.
2.  **Common Oxidation State:** The most common and stable oxidation state for both series is +3. This is due to the loss of two s-electrons and one d- or f-electron.
3.  **Lanthanoid/Actinoid Contraction:** Both series exhibit a gradual decrease in atomic and ionic radii (for M^3+ ions) across the series. This is known as lanthanoid contraction and actinoid contraction, respectively, caused by the poor shielding effect of f-electrons.
4.  **Electropositivity:** Both are highly electropositive metals and are very reactive, especially with non-metals.
5.  **Reactivity:** They are generally reactive metals, reacting with water and acids to liberate hydrogen.
6.  **Complex Formation:** Both show a tendency to form complexes, though actinoids have a greater tendency due to larger size and more variable oxidation states.
7.  **Magnetic Properties:** Many ions of both series are paramagnetic due to the presence of unpaired f-electrons.
8.  **Spectra:** Both exhibit characteristic absorption spectra due to f-f transitions.

#### Differences:

In summary, the key differences arise from the fact that 5f orbitals in actinoids are less effectively shielded and more exposed than 4f orbitals in lanthanoids. This leads to a greater range of oxidation states, more pronounced contraction, and higher reactivity in actinoids.

Question 27

Q27. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Accepted Answer

### Lanthanoid Contraction
Lanthanoid contraction refers to the gradual decrease in atomic and ionic radii of the lanthanoid elements (from La to Lu) as the atomic number increases. This contraction is observed across the 4f series.

**Cause:**
The primary cause of lanthanoid contraction is the poor shielding effect of the 4f electrons. As we move across the lanthanoid series, the nuclear charge increases by one unit at each successive element. The added electron enters a 4f orbital. However, 4f orbitals have a very diffuse shape and are not very effective at shielding the outer electrons from the increasing nuclear charge. Consequently, the effective nuclear charge experienced by the outer electrons increases significantly, pulling them closer to the nucleus and resulting in a steady decrease in atomic and ionic radii.

### Consequences of Lanthanoid Contraction
Lanthanoid contraction has several significant consequences, particularly for the elements following the lanthanoids in the periodic table (i.e., the 5d transition series):

1.  **Similarity in Atomic Radii of 4d and 5d Transition Elements:** Due to lanthanoid contraction, the atomic radii of elements in the 5d transition series are very similar to those of their corresponding elements in the 4d transition series. For example, the atomic radius of Zr (4d) is 160 pm, and that of Hf (5d) is 159 pm. Similarly, Nb (4d) and Ta (5d) have very similar radii. This similarity in size leads to very similar chemical properties for these pairs of elements, making their separation difficult.

2.  **Higher Density of 5d Transition Elements:** Because the atomic masses of 5d elements are significantly higher than those of 4d elements, but their atomic volumes are nearly the same (due to similar radii), the 5d transition elements have much higher densities than their 4d counterparts.

3.  **Basicity of Lanthanoid Hydroxides:** As the size of the lanthanoid ions (Ln³⁺) decreases from La³⁺ to Lu³⁺ due to lanthanoid contraction, the covalent character of the Ln–OH bond increases (Fajan's rule), and the basicity of their hydroxides (Ln(OH)₃) decreases. Thus, La(OH)₃ is the most basic, and Lu(OH)₃ is the least basic.

4.  **Difficulty in Separation of Lanthanoids:** The small difference in ionic radii between successive lanthanoids makes their chemical properties very similar. This similarity makes their separation from each other quite challenging, often requiring advanced techniques like ion-exchange chromatography.

5.  **Effect on Post-Lanthanoid Elements:** Elements immediately following the lanthanoids (e.g., Hf, Ta, W, Re, Os, Ir, Pt, Au, Hg) exhibit properties that are significantly influenced by the smaller-than-expected atomic radii caused by lanthanoid contraction. This affects their metallic character, ionization energies, and complex-forming abilities.

Question 28

Q28. What are the characteristics of the interstitial compounds?

Accepted Answer

Interstitial compounds are formed when small atoms like hydrogen, carbon, nitrogen, or boron get trapped in the interstitial voids (spaces) within the crystal lattice of transition metals. These compounds are typically non-stoichiometric and possess distinct properties from their parent metals.

Here are the main characteristics of interstitial compounds:

1.  **High Melting Points:** They generally have much higher melting points than the pure metals from which they are formed. This is due to the strong covalent bonds formed between the metal atoms and the interstitial atoms.
2.  **Extreme Hardness:** They are typically very hard and rigid, often approaching the hardness of diamond. For example, some borides and carbides of transition metals are extremely hard.
3.  **Retain Metallic Conductivity:** Despite the presence of non-metal atoms, they usually retain the metallic conductivity of their parent metals. This is because the valence electrons of the metal atoms are still delocalized.
4.  **Chemical Inertness:** They are chemically very inert and unreactive. This stability arises from the strong bonding within their lattice structure.
5.  **Non-stoichiometric Nature:** Many interstitial compounds are non-stoichiometric, meaning they do not have a fixed integer ratio of atoms (e.g., TiH₁.₇, Fe₃C, VH₀.₅₆). This is because the small atoms can occupy a variable number of interstitial sites.
6.  **Ferromagnetism (in some cases):** While some metals are ferromagnetic, the formation of interstitial compounds can sometimes enhance or diminish this property. For example, Fe₃C (cementite) is ferromagnetic.
7.  **Catalytic Activity:** Many interstitial compounds, particularly carbides and nitrides, exhibit good catalytic activity, making them useful in various industrial processes.

Question 29

Q29. How is the variability in oxidation states of transition metals different from that of the non-transition metals?

Accepted Answer

The variability in oxidation states is a characteristic feature of both transition metals and non-transition metals (main group elements), but there are fundamental differences in the reasons for this variability and the nature of the oxidation states exhibited.

### Variability in Oxidation States of Transition Metals

1.  **Origin of Variability:** The variability in oxidation states of transition metals arises primarily from the involvement of both (n-1)d and ns electrons in bonding. The energy difference between the (n-1)d and ns orbitals is very small, allowing electrons from both sets of orbitals to participate in chemical bonding.
2.  **Small Increments:** Transition metals typically exhibit oxidation states that differ by unity (e.g., Fe can show +2, +3; Mn can show +2, +3, +4, +5, +6, +7). This is because individual d-electrons can be successively removed or shared.
3.  **Common Oxidation States:** The most common oxidation states for transition metals are usually +2 and +3, but higher oxidation states are also observed, especially in the middle of the series (e.g., Mn shows +7).
4.  **Stability:** Lower oxidation states are often found in ionic compounds, while higher oxidation states are more stable when the metal is bonded to highly electronegative elements like oxygen or fluorine, often forming covalent compounds (e.g., Mn₂O₇, CrO₃).
5.  **Redox Properties:** The ability to exhibit multiple stable oxidation states makes transition metals and their compounds excellent catalysts and redox reagents.

### Variability in Oxidation States of Non-Transition Metals (Main Group Elements)

1.  **Origin of Variability:** For non-transition metals, the variability in oxidation states primarily arises from the involvement of only the valence s and p electrons. The energy difference between s and p orbitals within the same shell is generally larger than between (n-1)d and ns orbitals in transition metals.
2.  **Larger Increments:** The oxidation states of non-transition metals often differ by two units (e.g., Group 13: +1, +3; Group 14: +2, +4; Group 15: +3, +5). This is due to the 'inert pair effect' in heavier elements, where the ns² electrons become reluctant to participate in bonding, leading to lower oxidation states.
3.  **Common Oxidation States:** The most common oxidation states are usually determined by their group number (e.g., Group 1: +1, Group 2: +2, Group 13: +3, Group 14: +4, Group 15: +3 or +5, Group 16: -2 or +2/+4/+6, Group 17: -1 or +1/+3/+5/+7).
4.  **Stability:** For heavier elements in groups 13-16, the lower oxidation state (e.g., +1 for Tl, +2 for Pb, +3 for Bi) becomes more stable due to the inert pair effect, while the higher oxidation state (group oxidation state) becomes less stable.
5.  **Limited Redox Properties:** While some non-transition elements exhibit redox properties, their range and versatility are generally less extensive than those of transition metals.

### Key Differences Summarized:

In essence, the availability of closely spaced (n-1)d and ns orbitals allows transition metals to exhibit a rich and continuous variation in oxidation states, differing by single units, which is distinct from the more discrete and often two-unit differences observed in non-transition metals, where the inert pair effect plays a significant role for heavier elements.

Question 30

Q30. Describe the preparation of potassium dichromate from chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Accepted Answer

### Preparation of Potassium Dichromate (K₂Cr₂O₇) from Chromite Ore (FeCr₂O₄)
Potassium dichromate is prepared from chromite ore (FeCr₂O₄) in a three-step process:

**Step 1: Formation of Sodium Chromate (Na₂CrO₄)**
Finely powdered chromite ore is mixed with sodium carbonate (Na₂CO₃) or potassium carbonate (K₂CO₃) and lime (CaO) and roasted in a reverberatory furnace in the presence of excess air (oxygen) at a high temperature (around 1000-1200 °C). The iron present in the ore is oxidized to ferric oxide, and chromium is oxidized to sodium chromate.

4FeCr₂O₄ (s) + 8Na₂CO₃ (s) + 7O₂ (g) → 8Na₂CrO₄ (s) + 2Fe₂O₃ (s) + 8CO₂ (g)

The roasted mass is then extracted with water. Sodium chromate (Na₂CrO₄) is soluble in water, while ferric oxide (Fe₂O₃) is insoluble and can be filtered off.

**Step 2: Conversion of Sodium Chromate to Sodium Dichromate (Na₂Cr₂O₇)**
The yellow solution of sodium chromate is acidified with sulfuric acid (H₂SO₄). Sodium chromate, being basic, reacts with the acid to form sodium dichromate. This reaction involves the condensation of two chromate ions.

2Na₂CrO₄ (aq) + H₂SO₄ (aq) → Na₂Cr₂O₇ (aq) + Na₂SO₄ (aq) + H₂O (l)

Upon concentration, sodium dichromate (Na₂Cr₂O₇) crystallizes out. Sodium dichromate is more soluble than potassium dichromate, which is why it's often converted to potassium dichromate in the next step.

**Step 3: Conversion of Sodium Dichromate to Potassium Dichromate (K₂Cr₂O₇)**
Sodium dichromate solution is treated with potassium chloride (KCl). Due to the lower solubility of potassium dichromate compared to sodium dichromate, potassium dichromate precipitates out as orange-red crystals upon cooling.

Na₂Cr₂O₇ (aq) + 2KCl (aq) → K₂Cr₂O₇ (s) + 2NaCl (aq)

Potassium dichromate is then separated by filtration and purified by recrystallization.

### Effect of Increasing pH on a Solution of Potassium Dichromate
Potassium dichromate (K₂Cr₂O₇) exists in an equilibrium with potassium chromate (K₂CrO₄) in aqueous solution. This equilibrium is highly dependent on the pH of the solution.

The equilibrium reaction is:

Cr₂O₇²⁻ (aq) (Orange) + H₂O (l) ⇌ 2CrO₄²⁻ (aq) (Yellow) + 2H⁺ (aq)

*   **Increasing pH (making the solution alkaline/basic):** If the pH of a potassium dichromate solution is increased (i.e., H⁺ ions are removed by adding a base like NaOH), the equilibrium will shift to the **right** according to Le Chatelier's principle. This means that the orange dichromate ions (Cr₂O₇²⁻) will be converted into yellow chromate ions (CrO₄²⁻). Therefore, the solution will change color from **orange to yellow**.

Cr₂O₇²⁻ (aq) (Orange) + 2OH⁻ (aq) → 2CrO₄²⁻ (aq) (Yellow) + H₂O (l)

*   **Decreasing pH (making the solution acidic):** Conversely, if the pH is decreased (i.e., H⁺ ions are added by adding an acid), the equilibrium will shift to the **left**, converting yellow chromate ions back to orange dichromate ions. The solution will change color from **yellow to orange**.

This interconversion between chromate and dichromate ions is a classic example of a pH-dependent equilibrium in transition metal chemistry.

Question 31

Q31. What is the effect of lanthanoid contraction on the chemistry of the elements of third transition series?

Accepted Answer

Lanthanoid contraction is the gradual decrease in atomic and ionic radii of the lanthanoids (from La to Lu) due to the poor shielding effect of 4f electrons. This contraction has significant consequences on the chemistry of the elements of the third (5d) transition series, particularly when compared to the elements of the second (4d) transition series.

**Effects of Lanthanoid Contraction on the Chemistry of Third Transition Series Elements:**

1.  **Similar Atomic and Ionic Radii:** Due to lanthanoid contraction, the elements of the third transition series (5d series) have atomic and ionic radii very similar to their corresponding elements in the second transition series (4d series). For example, the atomic radius of Zr (4d) is 160 pm and Hf (5d) is 159 pm; Nb (4d) is 147 pm and Ta (5d) is 143 pm. This similarity in size is a direct consequence of the contraction, as the expected increase in size from 4d to 5d series is nullified.

2.  **Similar Chemical Properties:** Because of their similar atomic and ionic radii and similar outer electronic configurations, the elements of the 4d and 5d transition series (especially those in the same group, e.g., Zr-Hf, Nb-Ta, Mo-W) exhibit very similar chemical properties. This makes their separation in nature quite difficult.

3.  **Higher Ionization Enthalpies:** The increased nuclear charge and smaller atomic size of the 5d elements (due to lanthanoid contraction) lead to stronger attraction of the valence electrons towards the nucleus. Consequently, the 5d elements have significantly higher ionization enthalpies compared to their 4d counterparts. This makes them less reactive than 4d elements in some cases.

4.  **Higher Densities:** The smaller atomic volume combined with a relatively high atomic mass for the 5d elements results in very high densities for these elements. For example, Hf (density = 13.31 g/cm³) has a much higher density than Zr (density = 6.51 g/cm³), and Os (density = 22.59 g/cm³) and Ir (density = 22.56 g/cm³) are among the densest known elements.

5.  **Increased Electronegativity:** The smaller size and increased effective nuclear charge also lead to a slight increase in electronegativity for the 5d elements compared to their 4d counterparts.

In summary, lanthanoid contraction is a crucial factor that dictates the physical and chemical properties of the third transition series elements, making them remarkably similar to their second transition series congeners and distinct from what would be expected based on simple periodic trends.

Question 32

Q32. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

Accepted Answer

Cerium (Ce) is a member of the lanthanoid series which is well known to exhibit a stable +4 oxidation state. Its electronic configuration is [Xe] 4f¹ 5d¹ 6s². In the +4 oxidation state, it attains a stable noble gas configuration ([Xe]), making Ce⁴⁺ a strong oxidizing agent.

Question 33

Q33. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Accepted Answer

Actinoid contraction is greater from element to element than lanthanoid contraction primarily due to the following reasons:

1.  **Poorer Shielding by 5f Electrons:** The 5f orbitals in actinoids have a more diffuse shape and are less effective at shielding the outer electrons from the nuclear charge compared to the 4f orbitals in lanthanoids. This means that as the atomic number increases across the actinoid series, the effective nuclear charge experienced by the outer electrons increases more significantly, leading to a stronger pull and a greater contraction.

2.  **Larger Number of Electrons in 5f Orbitals:** The 5f orbitals begin filling at the start of the actinoid series, and there are 14 electrons that can occupy these orbitals. The cumulative effect of the poor shielding by these 5f electrons over 14 elements leads to a more pronounced contraction compared to the 4f electrons.

3.  **Relativistic Effects:** Relativistic effects, which become more significant for heavier elements, also contribute to the greater contraction in actinoids. These effects influence the energies and sizes of orbitals, particularly for inner electrons, indirectly affecting the overall atomic size.

Question 34

Q34. Discuss the general characteristics of lanthanoids with special reference to their electronic configuration, oxidation states, chemical reactivity and comparison with actinoids.

Accepted Answer

The lanthanoids (or lanthanides) are a series of 14 elements from Cerium (Ce, Z=58) to Lutetium (Lu, Z=71) that follow Lanthanum (La, Z=57) in the periodic table. They are characterized by the filling of the 4f orbitals. Their general characteristics are discussed below:

### I. Electronic Configuration

The general electronic configuration of lanthanoids is [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s². The filling of 4f orbitals is not perfectly regular due to the comparable energies of 4f, 5d, and 6s orbitals. The most common configurations have 5d⁰, with the exception of Gadolinium (Gd) and Lutetium (Lu).

*   **Lanthanum (La, Z=57):** [Xe] 5d¹ 6s² (not a lanthanoid, but its configuration sets the stage)
*   **Cerium (Ce, Z=58):** [Xe] 4f¹ 5d¹ 6s² (or [Xe] 4f² 6s²)
*   **Gadolinium (Gd, Z=64):** [Xe] 4f⁷ 5d¹ 6s² (due to stable half-filled 4f⁷ configuration)
*   **Lutetium (Lu, Z=71):** [Xe] 4f¹⁴ 5d¹ 6s² (due to stable completely filled 4f¹⁴ configuration)

### II. Oxidation States

The most common and stable oxidation state for all lanthanoids is +3. This is because the sum of the first three ionization enthalpies is relatively low, and the removal of three electrons (two 6s and one 5d or 4f) leads to stable configurations. However, some lanthanoids also exhibit +2 and +4 oxidation states, especially when these lead to a stable empty (4f⁰), half-filled (4f⁷), or completely filled (4f¹⁴) f-subshell.

*   **+2 Oxidation State:** Observed in Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴). Sm²⁺ (4f⁶) and Tm²⁺ (4f¹³) also exist but are less stable.
*   **+4 Oxidation State:** Observed in Ce⁴⁺ (4f⁰) and Tb⁴⁺ (4f⁷). Pr⁴⁺ (4f¹) and Nd⁴⁺ (4f²) also exist but are less stable.

### III. Chemical Reactivity

Lanthanoids are highly electropositive metals and are quite reactive, similar to alkaline earth metals. Their reactivity generally decreases across the series due to lanthanoid contraction.

*   **Reaction with Air:** They tarnish rapidly in air, forming oxides (Ln₂O₃). Cerium forms CeO₂.
*   **Reaction with Water:** They react slowly with cold water and rapidly with hot water to liberate hydrogen, forming hydroxides (Ln(OH)₃).
*   **Reaction with Acids:** They readily react with dilute acids (e.g., HCl) to liberate hydrogen.
*   **Reaction with Halogens:** They combine with halogens (X₂) to form trihalides (LnX₃).
*   **Reaction with Sulfur:** They react with sulfur to form sulfides (Ln₂S₃).
*   **Reaction with Carbon:** They form carbides (e.g., Ln₃C, Ln₂C₃, LnC₂).

### IV. Comparison with Actinoids

In summary, while both lanthanoids and actinoids are f-block elements, actinoids show greater variability in oxidation states, are more radioactive, and exhibit a more pronounced contraction due to the more diffuse and less effective shielding of 5f electrons.

Question 35

Q35. What is the effect of lanthanoid contraction on the chemistry of the elements of third transition series?

Accepted Answer

Lanthanoid contraction is the gradual decrease in the atomic and ionic radii of lanthanoids (from La to Lu) as the atomic number increases. This contraction is a consequence of the poor shielding effect of the 4f electrons. The 4f orbitals are deeply embedded and do not effectively shield the outer electrons from the increasing nuclear charge. As a result, the effective nuclear charge experienced by the outer electrons increases, pulling them closer to the nucleus and causing a decrease in size.

**Effects of Lanthanoid Contraction on the Chemistry of Third Transition Series Elements (5d Series):**

1.  **Similar Atomic and Ionic Radii of 4d and 5d Series Elements:**
    Due to lanthanoid contraction, the elements of the third transition series (5d series) have atomic and ionic radii very similar to those of the corresponding elements in the second transition series (4d series). For example, the atomic radius of Zirconium (Zr, 4d series) is 160 pm, and that of Hafnium (Hf, 5d series) is 159 pm. Similarly, Niobium (Nb, 4d) and Tantalum (Ta, 5d) have very similar radii. This similarity in size is the most significant consequence.

2.  **Increased Density of 5d Series Elements:**
    Because the atomic mass increases significantly from the 4d to the 5d series, but the atomic volume remains nearly the same (due to similar radii), the density of the 5d transition elements is much higher than that of the 4d transition elements. For instance, the density of Hf is 13.3 g/cm³, while that of Zr is 6.51 g/cm³.

3.  **Similar Chemical Properties of 4d and 5d Series Elements:**
    The similar atomic and ionic radii, coupled with similar electronic configurations (especially in their common oxidation states), lead to very similar chemical properties for elements of the same group in the 4d and 5d transition series. This makes their separation difficult. For example, Zr and Hf are chemically very similar and are found together in minerals.

4.  **Higher Ionization Enthalpies of 5d Series Elements:**
    Due to the increased effective nuclear charge and smaller atomic size, the electrons in the 5d series elements are held more tightly. Consequently, their ionization enthalpies are generally higher than those of the corresponding 4d and 3d series elements. This contributes to the greater stability of higher oxidation states for some 5d elements.

5.  **Increased Electronegativity:**
    The increased effective nuclear charge also leads to a slight increase in electronegativity for 5d elements compared to their 4d counterparts.

Question 36

Q36. Discuss the general characteristics of lanthanoids with special reference to their electronic configuration, oxidation states, chemical reactivity and magnetic properties.

Accepted Answer

Lanthanoids are a series of 14 elements from Cerium (Ce, Z=58) to Lutetium (Lu, Z=71) that follow Lanthanum (La, Z=57) in the periodic table. They are characterized by the filling of 4f orbitals. They are also known as rare earth elements.

**1. Electronic Configuration:**
*   **General Electronic Configuration:** The general outer electronic configuration of lanthanoids is [Xe] 4f^(1-14) 5d^(0-1) 6s^2. The 4f orbitals are progressively filled. However, there are some irregularities.
*   **Irregularities:** The 5d orbital remains empty in most lanthanoids, but it contains one electron in Gadolinium (Gd, [Xe] 4f^7 5d^1 6s^2) and Lutetium (Lu, [Xe] 4f^14 5d^1 6s^2). This is due to the extra stability associated with half-filled (f^7) and completely filled (f^14) 4f subshells.
    *   La (Z=57): [Xe] 5d^1 6s^2
    *   Ce (Z=58): [Xe] 4f^1 5d^1 6s^2 (or sometimes written as 4f^2 6s^2)
    *   Gd (Z=64): [Xe] 4f^7 5d^1 6s^2
    *   Lu (Z=71): [Xe] 4f^14 5d^1 6s^2

**2. Oxidation States:**
*   **Most Common Oxidation State:** The most characteristic and stable oxidation state for all lanthanoids is **+3**. This is because the sum of the first three ionization enthalpies is relatively low, allowing for the removal of two 6s electrons and one 5d or 4f electron.
*   **Other Oxidation States (+2 and +4):** Some lanthanoids also exhibit +2 and +4 oxidation states, particularly when these states lead to stable f^0, f^7, or f^14 configurations.
    *   **+2 Oxidation State:** Observed in Europium (Eu^2+, [Xe] 4f^7) and Ytterbium (Yb^2+, [Xe] 4f^14). These ions are relatively stable due to their half-filled and completely filled f-subshells, respectively. Samarium (Sm^2+) and Thulium (Tm^2+) also exist but are less stable and act as strong reducing agents.
    *   **+4 Oxidation State:** Observed in Cerium (Ce^4+, [Xe] 4f^0) and Terbium (Tb^4+, [Xe] 4f^7). Ce^4+ is a strong oxidizing agent as it readily reverts to the more stable Ce^3+ state. Praseodymium (Pr^4+) and Neodymium (Nd^4+) also exist but are less stable and act as strong oxidizing agents.

**3. Chemical Reactivity:**
Lanthanoids are highly electropositive metals and are quite reactive, similar to alkaline earth metals, but generally less reactive than alkali metals.
*   **Reaction with Air/Oxygen:** They tarnish rapidly in air, forming a protective oxide layer. They burn in oxygen at high temperatures to form M₂O₃ type oxides (e.g., 4M + 3O₂ → 2M₂O₃). Cerium forms CeO₂.
*   **Reaction with Water:** They react slowly with cold water and rapidly with hot water to liberate hydrogen, forming hydroxides (2M + 6H₂O → 2M(OH)₃ + 3H₂).
*   **Reaction with Acids:** They readily react with dilute acids (e.g., HCl) to liberate hydrogen (2M + 6HCl → 2MCl₃ + 3H₂).
*   **Reaction with Halogens:** They react with halogens (X₂) at elevated temperatures to form trihalides (2M + 3X₂ → 2MX₃).
*   **Reaction with Sulfur:** They react with sulfur to form sulfides (2M + 3S → M₂S₃).
*   **Reaction with Carbon:** They form carbides (e.g., MC₂, M₃C, M₂C₃) when heated with carbon.
*   **Reaction with Nitrogen:** They form nitrides (MN) at high temperatures.

**4. Magnetic Properties:**
*   **Paramagnetism:** Most lanthanoid ions (M^3+) are paramagnetic because they contain unpaired electrons in their 4f orbitals. The degree of paramagnetism varies with the number of unpaired electrons.
*   **Diamagnetism:** La^3+ ([Xe] 4f^0) and Lu^3+ ([Xe] 4f^14) ions are diamagnetic because they have no unpaired electrons.
*   **Orbital Contribution:** Unlike transition metals (d-block elements) where the magnetic moment is primarily due to spin, for lanthanoids, the orbital motion of electrons also contributes significantly to the total magnetic moment. The 4f orbitals are deeply shielded by 5s and 5p electrons, so they are not significantly affected by the crystal field of the ligands. Therefore, the orbital angular momentum is not quenched, and the magnetic moment is calculated using the formula involving total angular momentum (J):
    µ = g√(J(J+1)) Bohr magnetons (where g is the Lande g-factor).
*   The magnetic moments of lanthanoid ions are generally higher than those of d-block elements with a similar number of unpaired electrons.

Chapter 8: The d and f Block Elements