Block D and F Elements

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.

8.1 The elements which have partially filled d or f subshells in any common oxidation state are called as the transition elements. Silver (the Atomic number is 47 and electronic configuration is [Kr] 4d¹⁰5s¹) has a completely filled 4d orbital in its ground state but has two oxidation states (+1, +2).

In the +1 oxidation state, an electron is removed from the s-orbital and in +2 oxidation state, one electron from d-orbital is also removed. Thus, the d-orbital now becomes partially filed (4d⁹). Hence, it is a transition element.

Students can refer the NCERT Solutions for Class 12 Chemistry Chapter 4 The d and f block Elements shared by the experts at Shiksha for NCERT Chemisytry textbook questions. The solutions for CBSE Chemistry d and f block questions is explanined in easy way for better understanding.

Students need to practice the Class 12 Chemistry previous year question to identify the important questions. Some of important questions are reactions of $K_2Cr_2O_7$K2? Cr2? O7? and $KMnO_4$KMnO4? in acidic, basic, and neutral medium, lanthanide contraction and its consequences, balancing redox reactions, etc.

Students can use the class 12 previous year d and f block questions pdf to prepare for the exam and self-assessment. NCERT solutions enhace the proble sloving skills by explaning the complex questions in easy way.