CBSE Class 12 Chemistry Chapter 8: The d and f Block Elements - NCERT Solutions

Solution: To determine the electronic configuration of ions, we first write the electronic configuration of the neutral atom and then remove the electrons from the outermost shells.

1. Cr³⁺: The electronic configuration of neutral Chromium (Cr) is [Ar] 3d⁵ 4s¹. To form Cr³⁺, we remove one electron from the 4s orbital and two electrons from the 3d orbital. Thus, the configuration is [Ar] 3d³.

2. Pm³⁺: Promethium (Pm) has atomic number 61. Its electronic configuration is [Xe] 4f⁵ 6s². To form Pm³⁺, we remove two electrons from the 6s orbital and one electron from the 4f orbital. Thus, the configuration is [Xe] 4f⁴.

3. Cu⁺: The electronic configuration of neutral Copper (Cu) is [Ar] 3d¹⁰ 4s¹. To form Cu⁺, we remove one electron from the 4s orbital. Thus, the configuration is [Ar] 3d¹⁰. This is a stable, fully-filled d-orbital configuration.

4. Ce⁴⁺: Cerium (Ce) has atomic number 58. Its electronic configuration is [Xe] 4f¹ 5d¹ 6s². To form Ce⁴⁺, we remove all four valence electrons. Thus, the configuration is [Xe]. This is a stable noble gas configuration.

5. Co²⁺: The electronic configuration of neutral Cobalt (Co) is [Ar] 3d⁷ 4s². To form Co²⁺, we remove two electrons from the 4s orbital. Thus, the configuration is [Ar] 3d⁷.

6. Lu²⁺: Lutetium (Lu) has atomic number 71. Its electronic configuration is [Xe] 4f¹⁴ 5d¹ 6s². To form Lu²⁺, we remove two electrons from the 6s orbital. Thus, the configuration is [Xe] 4f¹⁴ 5d¹.

7. Mn²⁺: The electronic configuration of neutral Manganese (Mn) is [Ar] 3d⁵ 4s². To form Mn²⁺, we remove two electrons from the 4s orbital. Thus, the configuration is [Ar] 3d⁵. This is a stable, half-filled d-orbital configuration.

8. Th⁴⁺: Thorium (Th) has atomic number 90. Its electronic configuration is [Rn] 6d² 7s². To form Th⁴⁺, we remove two electrons from the 7s orbital and two electrons from the 6d orbital. Thus, the configuration is [Rn]. This is a stable noble gas configuration.

Solution: The stability of an ion is often related to its electronic configuration, particularly the stability of its d-orbitals. Half-filled and fully-filled orbitals possess extra stability due to symmetrical electron distribution and reduced electron-electron repulsion.

The electronic configuration of Mn²⁺ is [Ar] 3d⁵. This configuration features a half-filled d-subshell, which is particularly stable.

The electronic configuration of Fe²⁺ is [Ar] 3d⁶. To achieve the +3 oxidation state, Fe²⁺ needs to lose one more electron, resulting in the configuration [Ar] 3d⁵.

Since Mn²⁺ already possesses the stable 3d⁵ configuration, it requires significantly more energy to oxidize it to Mn³⁺ ([Ar] 3d⁴). In contrast, Fe²⁺ readily loses one electron to form Fe³⁺ ([Ar] 3d⁵) because the resulting configuration is more stable than the initial 3d⁶ configuration. Therefore, Mn²⁺ compounds are more stable than Fe²⁺ compounds towards oxidation to the +3 state.

Solution: In the first half of the first row of transition elements (from Sc to Mn), the +2 oxidation state generally becomes more stable as the atomic number increases. This trend can be understood by examining the electronic configurations of these elements and their +2 ions.

The first-row transition elements are Sc (Z=21) to Mn (Z=25). Their neutral electronic configurations are:

• Sc: [Ar] 3d¹ 4s²
• Ti: [Ar] 3d² 4s²
• V: [Ar] 3d³ 4s²
• Cr: [Ar] 3d⁵ 4s¹
• Mn: [Ar] 3d⁵ 4s²

When these elements form +2 ions, they lose the two 4s electrons:

• Sc²⁺: [Ar] 3d¹
• Ti²⁺: [Ar] 3d²
• V²⁺: [Ar] 3d³
• Cr²⁺: [Ar] 3d⁴
• Mn²⁺: [Ar] 3d⁵

As we move from Sc to Mn, the number of electrons in the 3d subshell increases from 1 to 5. The stability of the d-subshell increases as it becomes more filled. The half-filled d⁵ configuration of Mn²⁺ is particularly stable. Therefore, the +2 oxidation state becomes progressively more stable from Sc²⁺ to Mn²⁺ because the resulting d-orbital configurations are increasingly stable due to the filling of the d-subshell.