Question

Which from following pair of elements in their respective oxidation states have same number of unpaired electrons ?

• A. \(\text{Fe}^{2+}\) and \(\text{Mn}^{2+}\)
• B. \(\text{Co}^{2+}\) and \(\text{Ni}^{2+}\)
• C. \(\text{Fe}^{2+}\) and \(\text{Cr}^{2+}\)
• D. \(\text{Co}^{2+}\) and \(\text{Fe}^{2+}\)

Hint:

Write the 3d configuration of each ion and count unpaired electrons: Fe2+ and Cr2+ both have 4.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The number of unpaired electrons in a transition metal ion comes from its d-electron configuration, filled by Hund's rule (high spin for free ions).

Step 2: Key Formula or Approach:
Atomic configurations: Cr \(3d^54s^1\), Mn \(3d^54s^2\), Fe \(3d^64s^2\), Co \(3d^74s^2\), Ni \(3d^84s^2\). For +2 ions the two 4s electrons are lost.

Step 3: Detailed Explanation:
\(\text{Cr}^{2+}\): \(3d^4\), so 4 unpaired electrons.
\(\text{Mn}^{2+}\): \(3d^5\), so 5 unpaired electrons.
\(\text{Fe}^{2+}\): \(3d^6\), one orbital paired and 4 unpaired electrons.
\(\text{Co}^{2+}\): \(3d^7\), so 3 unpaired electrons.
\(\text{Ni}^{2+}\): \(3d^8\), so 2 unpaired electrons.
Compare: (A) 4 and 5, (B) 3 and 2, (C) 4 and 4, (D) 3 and 4. Only (C) has equal numbers.

Final Answer:
\(\text{Fe}^{2+}\) and \(\text{Cr}^{2+}\) both have 4 unpaired electrons, option (C). \[ \boxed{\text{Fe}^{2+} \text{ and } \text{Cr}^{2+} \text{ (C)}} \]