Q26. Match List I with List II
List - I
A. Lanthanum
B. Praseodymium
C. Neodymium
D. Samarium
List - II
I. +3, +4
II. +3
III. +2, +3, +4
IV. +2, +3
Choose the correct answer from the options given bel

Question

Q26. Match List I with List II
List - I
A. Lanthanum
B. Praseodymium
C. Neodymium
D. Samarium
List - II
I. +3, +4
II. +3
III. +2, +3, +4
IV. +2, +3
Choose the correct answer from the options given below:
(a) A-II, B-IV, C-I, D-III
(b) A-IV, B-III, C-I, D-II
(c) A-II, B-I, C-III, D-IV
(d) A-IV, B-I, C-II, D-III

FirstInTest · Accepted Answer

Correct answer: c. A. In compounds, lanthanum exhibits only one oxidation state, +3. (A-II)B. Like most rare-earth elements, praseodymium most readily forms the +3 oxidation state, which is the only stable state in aqueous solution, although the +4 oxidation state is known in some solid compounds. (B-I)C. We know that, +3 oxidation is common in all series. Stable of +3 O.S. depends upon high hydration energy, high lactic energy and high I.P. Nd2+- has configuration 4f4. Nd4+- has configuration 4f2. Hence +2, +4 oxidation states are unstable in Nd in aqueous solution. (C-III)D. Samarium is a chemical element with symbol Sm and atomic number 62. It is a moderately hard silvery metal that slowly oxidizes in air. Being a typical member of the lanthanide series, samarium usually has the oxidation state +3. (D-IV)So matches are: A-II, B-I, C-III, D-IV. This matches option (c).

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