Practicing Success
Match List I with
Choose the correct answer from the options given below: |
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 |
A-II, B-I, C-III, D-IV |
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. |