Question

Consider \(|x|\) is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with \(|x|\) number of unpaired electrons from the following are:
A. \(Sc^{3+}\)
B. \(Zn^{2+}\)
C. \(V^{2+}\)
D. \(Fe^{2+}\)
E. \(Co^{2+}\)
Choose the correct answer from the options given below:

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

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We first identify the highest oxidation states of Manganese in its compounds and find the difference \(|x|\). Then, we determine the number of unpaired electrons in the given transition metal ions to find those that match \(|x|\).

Step 2: Key Formula or Approach:
1. Highest Manganese Fluoride: \(MnF_4\) (Oxidation state of Mn = +4).
2. Highest Manganese Oxide: \(Mn_2O_7\) (Oxidation state of Mn = +7).
3. \(|x| = |7 - 4| = 3\).

Step 3: Detailed Explanation:
We are looking for ions with 3 unpaired electrons (\(n = 3\)):
A. \(Sc^{3+}\): \([Ar] 3d^0\). Unpaired electrons = 0.
B. \(Zn^{2+}\): \([Ar] 3d^{10}\). Unpaired electrons = 0.
C. \(V^{2+}\): \([Ar] 3d^3\). Unpaired electrons = 3. (Matches)
D. \(Fe^{2+}\): \([Ar] 3d^6\). Unpaired electrons = 4.
E. \(Co^{2+}\): \([Ar] 3d^7\). In a typical high-spin state, \(t_{2g}^5 e_g^2\), unpaired electrons = 3. (Matches)
Thus, C and E satisfy the condition.

Step 4: Final Answer:
The ions are C and E only.