Question

Consider the given central metal ions of low spin complex and choose the correct increasing order of unpaired electrons: Mn\(^{+} \), Cr\(^{+} \), Fe\(^{+} \), Co\(^{+} \).

• A. Co\(^{+} \)<Fe\(^{+} \)<Mn\(^{+} \)<Cr\(^{+} \)
• B. Co\(^{+} \)<Mn\(^{+} \)<Fe\(^{+} \)<Cr\(^{+} \)
• C. Cr\(^{+} \)<Mn\(^{+} \)<Cr\(^{+} \)<Fe\(^{+} \)
• D. Cr\(^{+} \)<Mn\(^{+} \)<Co\(^{+} \)<Fe\(^{+} \)

Hint:

In transition metal complexes, the number of unpaired electrons can be determined from the electron configuration of the metal ion.

Answer 1

The Correct Option is A

Step 1: Electron configuration.
The electron configurations of the central metal ions in their respective oxidation states determine the number of unpaired electrons. The order of unpaired electrons follows the general trend: \[ \text{Co}^{+}<\text{Fe}^{+}<\text{Mn}^{+}<\text{Cr}^{+} \] Step 2: Conclusion.
Thus, the correct order of increasing unpaired electrons is Co\(^{+} \)<Fe\(^{+} \)<Mn\(^{+} \)<Cr\(^{+} \). Final Answer: \[ \boxed{\text{Co}^{+}<\text{Fe}^{+}<\text{Mn}^{+}<\text{Cr}^{+}} \]