Question

If \(|x|\) represents the difference in maximum oxidation state possible for Mn in its oxide and in its fluoride. Among the following ions, which ions contain unpaired electrons equal to \(|x|\). (A) \(Zn^{2+}\) (B) \(Fe^{2+}\) (C) \(Co^{2+}\) (D) \(V^{2+}\) (E) \(Sc^{3+}\)

• A. A and E
• B. C and D
• C. B and C
• D. B and D

Answer 1

The Correct Option is B

To solve this problem, we need to find the difference in the maximum oxidation state of Mn in its oxide and fluoride compounds and determine which ions have unpaired electrons equal to this difference.

1. Determine the maximum oxidation state of Mn in its oxide:
   • Manganese (Mn) can exhibit a wide range of oxidation states. In its highest oxidation state in oxides, Mn forms \( \text{MnO}_4^- \) where Mn is in the +7 oxidation state.
2. Determine the maximum oxidation state of Mn in its fluoride:
   • In the fluoride \( \text{MnF}_3 \), manganese is in the +3 oxidation state, which is the highest oxidation state in stable fluoride compounds.
3. Calculate the difference in oxidation states:
   • The difference between the maximum oxidation states of Mn in its oxide and fluoride is \( +7 \) (for oxide) - \( +3 \) (for fluoride) = \( 4 \).
4. Find ions with 4 unpaired electrons:
   • \(Zn^{2+}\): Electronic configuration is [Ar] \(3d^{10}\). No unpaired electrons.
   • \(Fe^{2+}\): Electronic configuration is [Ar] \(3d^{6}\). Has 4 unpaired electrons.
   • \(Co^{2+}\): Electronic configuration is [Ar] \(3d^{7}\). Has 3 unpaired electrons.
   • \(V^{2+}\): Electronic configuration is [Ar] \(3d^{3}\). Has 3 unpaired electrons.
   • \(Sc^{3+}\): Electronic configuration is [Ar] \(3d^{0}\). No unpaired electrons.

Hence, among the given ions, \( \text{Fe}^{2+} \) (with 4 unpaired electrons) matches the required criterion of having unpaired electrons equal to the calculated difference (|x| = 4).