Block D and F Elements

8.10 The 5f orbitals (in actinoids) have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in outer shells in case of actinoids is much more that experienced by lanthanoids.

As the effective nuclear charge experienced is high, the electrons are attracted with much force, hence the size of the atom decreases. Hence, actinoid contraction is greater from element to element than lanthanoid contraction.

8.9 The stability in aqueous condition depends on the hydration energy of the ions when they bond to the water molecules. And, the hydration energy is the amount of heat released as an ionic substance is dissolved and its constituent ions are hydrated or surrounded by water molecules.

Now, in Cu²⁺ and Cu⁺ ion, Cu²⁺ has a greater charge density than the Cu⁺ ion and so forms much stronger bonds releasing more energy. Therefore, in an aqueous medium, Cu²⁺ ion is more stable than Cu⁺ ion. This is because the energy required to remove one electron from Cu⁺ to Cu²⁺, is compensated by the high hydration energy of Cu²⁺.

8.8 Given, Z=atomic number = 27 = [Ar]3d⁷ 4s²

⇒ M²⁺ = [Ar]3d⁷

Hence, 3 unpaired electrons are present.

The spin only magnetic moment μ= √n(n+2)

Where n is the number of unpaired electrons.

Hence, μ =√(3*(3+2))

μ = √15 BM or μ = 3.87 BM

8.7 The following reactions are involved when Cr²⁺ and Fe²⁺ act as reducing

Cr²⁺ = Cr3+ ( E^? _{Cr3+/ Cr2+} = - 0.41 V)

Fe²⁺⇒ Fe³⁺ (E^? _{Fe3+/ Fe2+} = +0.77 V)

Since, Cr has less potential value, Cr²⁺ gets oxidised easily than Fe²⁺. Therefore, Cr²⁺ is a better reducing agent that Fe³⁺.

8.6 The oxidation state increases when an atom loses its Example: When Fe loses 2 electrons, its oxidation state becomes +2 from 0.

Oxygen (O) and fluorine (F) are very strong oxidizing agents. Both oxide and fluoride ions are highly electronegative and have a very small size, so they attract the electrons from metal atoms. Hence, they oxidize the metal to its the highest oxidation state.

8.4 The E⁰ (M²⁺/M) value of a metal depends on the energy changes involved in the following reactions:

1. Sublimation energy: The energy needed to convert one mole of atoms from a solid state to gaseous

2. Ionization energy: The energy supplied to remove electrons from one mole of atoms, which are in the gaseous

3. Hydration energy: The energy emitted to hydrate one mole of

Now, copper has a high ionisation energy and low hydration energy. Hence, the E⁰ (M²⁺/M) value for copper is positive.

8.3 Manganese (Z = 25) exhibits the largest number of oxidation states. This is because its electronic configuration is 3d⁵4s².

Because it has the maximum number of electrons (5 d electrons and 2s electrons) to easily lose and share.

Therefore, it can exhibit an oxidation state of +2 to +7. The compounds are as follows:

Mn (0) as Mn (s), Mn (II) as MnO, Mn (II, III), Mn etc.

8.2 The enthalpy of atomization depends on the strength of the metallic bonding. Stronger the metallic bonding, greater is the enthalpy of atomization. The metallic bonding is strong when there are more unpaired electrons in the atom.

All transition metals (except Zn, electronic configuration: [Kr] 3d¹⁰ 4s²), have at least one unpaired electron that is responsible for their stronger metallic bonding. Since the Zn atom does not have an unpaired electron, the metallic bonding is weak and hence the enthalpy of atomization is low.