Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: D and F Block Elements

Question:

Match List I with

List I List II
A. Lanthanum I. +3, +4
B. Praseodymium II. +3
C. Neodymium III. +2, +3, +4
D. Samarium IV. +2, +3

Choose the correct answer from the options given below:

Options:

A-II, B-IV, C-I, D-III

A-IV, B-III, C-I, D-II

A-II, B-I, C-III, D-IV

A-IV, B-I, C-II, D-III

Correct Answer:

A-II, B-I, C-III, D-IV

Explanation:

The correct answer is option 3. A-II, B-I, C-III, D-IV.

Let us look into the common oxidation states of each of the lanthanide elements mentioned:

A. Lanthanum (La):  Lanthanum is commonly found in the +3 oxidation state. In its +3 oxidation state, lanthanum loses three electrons to achieve a stable electron configuration. This is the most stable and common oxidation state for lanthanum in chemical compounds.

B. Praseodymium (Pr): Praseodymium exhibits several oxidation states, but the most common ones are +3 and +4. In its +3 oxidation state, praseodymium loses three electrons, similar to lanthanum. However, praseodymium can also exist in the +4 oxidation state, where it loses an additional electron, resulting in a higher positive charge.

C. Neodymium (Nd):  Neodymium, like praseodymium, can also exhibit multiple oxidation states. The most common ones are +2, +3, and +4. In its +3 oxidation state, neodymium loses three electrons, similar to lanthanum and praseodymium. However, neodymium can also exist in the +2 and +4 oxidation states, where it loses two or four electrons, respectively.

D. Samarium (Sm): Samarium, like neodymium, exhibits multiple oxidation states, with the most common ones being +2 and +3. In its +3 oxidation state, samarium loses three electrons, similar to lanthanum, praseodymium, and neodymium. It can also exist in the +2 oxidation state, where it loses two electrons.

In summary:

Lanthanum (La) commonly exhibits the +3 oxidation state.

Praseodymium (Pr) commonly exhibits the +3 and +4 oxidation states.

Neodymium (Nd) commonly exhibits the +2, +3, and +4 oxidation states.

Samarium (Sm) commonly exhibits the +2 and +3 oxidation states.

These oxidation states are based on the tendency of each lanthanide element to lose a certain number of electrons to achieve a stable electron configuration. The specific oxidation states observed in compounds of these elements can vary depending on factors such as the chemical environment and the nature of the bonding involved.