Match List I with List II List I List II A. Cu I. +3 B. Sc II. +7 C.…

The correct answer is option 2. A-IV, B-I, C-II, D-III.

List I	List II
A. Cu	IV. +1
B. Sc	I. +3
C. Mn	II. +7
D. Fe	III. +2

Let us go through each element in List I and explain their common oxidation states to match them with List II.

A.Copper (Cu)

Copper can exist in two common oxidation states:

+1: In compounds like copper(I) chloride (CuCl) and copper(I) oxide (Cu₂O), copper exhibits a +1 oxidation state.

+2: This is another common oxidation state, seen in compounds like copper(II) sulfate (CuSO₄).

However, in this question, we are looking for the +1 oxidation state for copper. This corresponds to IV in List II.

B. Scandium (Sc)

Scandium almost exclusively exhibits an oxidation state of +3 in all of its compounds. This is because scandium has a valence electron configuration of \( [Ar] 3d^1 4s^2 \). When it loses all three valence electrons, it achieves a stable noble gas configuration. Therefore, scandium is almost always found in the +3 oxidation state.

Examples:

Scandium chloride (ScCl₃)

Scandium oxide (Sc₂O₃)

Thus, +3 oxidation state corresponds to I in List II.

C. Manganese (Mn)

Manganese is a transition metal that can exhibit a wide range of oxidation states, including:

+2: Common in manganese(II) chloride (MnCl₂)

+4: Found in manganese dioxide (MnO₂)

+7: The highest oxidation state, found in compounds like potassium permanganate (KMnO₄).

In the +7 oxidation state, manganese is highly oxidizing, such as in the permanganate ion \( \text{MnO}_4^- \).

So, for this question, the +7 oxidation state corresponds to II in List II.

D. Iron (Fe)

Iron commonly exhibits two oxidation states:

+2: This occurs in compounds like iron(II) chloride (FeCl₂), where iron has lost two electrons.

+3: In compounds like iron(III) chloride (FeCl₃), iron has lost three electrons.

In this case, we are asked to find the +2 oxidation state, which corresponds to III in List II.