Match List I with List II for the oxidation state of central atoms:

Choose the correct answer from the options given below:

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

The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Here is a detailed explanation of how the oxidation states for each compound were determined, and how the correct matching is obtained:

(A) Cr₂O₇²⁻ (Dichromate ion):

The formula for the dichromate ion is Cr₂O₇²⁻.

Oxygen generally has an oxidation state of -2.

Let the oxidation state of chromium (Cr) be \( x \).

The total charge of the ion is -2.

The sum of the oxidation states must equal the total charge:

\(2x + 7(-2) = -2\)

\(2x - 14 = -2 ⇒ 2x = 12 ⇒ x = +6\)

So, the oxidation state of chromium (Cr) in Cr₂O₇²⁻ is +6.

(B) MnO₄⁻ (Permanganate ion):

The formula for the permanganate ion is MnO₄⁻.

Oxygen has an oxidation state of -2.

Let the oxidation state of manganese (Mn) be \( x \).

The total charge on the ion is -1.

The sum of the oxidation states must equal the charge on the ion:

\(x + 4(-2) = -1\)

\(x - 8 = -1 ⇒ x = +7\)

So, the oxidation state of manganese (Mn) in MnO₄⁻ is +7.

(C) VO₄³⁻ (Vanadate ion):

The formula for the vanadate ion is VO₄³⁻.

Oxygen has an oxidation state of -2.

Let the oxidation state of vanadium (V) be \( x \).

The total charge on the ion is -3.

The sum of the oxidation states must equal the charge on the ion:

\(x + 4(-2) = -3\)

\(x - 8 = -3 ⇒ x = +5\)

So, the oxidation state of vanadium (V) in VO₄³⁻ is +5.

(D) FeF₆³⁻ (Hexafluoroferrate ion):

The formula for the hexafluoroferrate ion is FeF₆³⁻.

Fluorine has an oxidation state of -1.

Let the oxidation state of iron (Fe) be \( x \).

The total charge on the ion is -3.

The sum of the oxidation states must equal the charge on the ion:

\(x + 6(-1) = -3\)

\(x - 6 = -3 ⇒ x = +3\)

So, the oxidation state of iron (Fe) in FeF₆³⁻ is +3.

The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)