Question

Consider the transition metal ions \( \text{Mn}^{3+}, \text{Cr}^{3+}, \text{Fe}^{3+} \) and \( \text{Co}^{3+} \) and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective \(d\)-orbitals of the complexes is

• A. \( \text{Cr}^{3+}>\text{Mn}^{3+}>\text{Fe}^{3+}>\text{Co}^{3+} \)
• B. \( \text{Fe}^{3+}>\text{Co}^{3+}>\text{Mn}^{3+}>\text{Cr}^{3+} \)
• C. \( \text{Mn}^{3+}>\text{Fe}^{3+}>\text{Co}^{3+}>\text{Cr}^{3+} \)
• D. \( \text{Cr}^{3+}>\text{Fe}^{3+}>\text{Co}^{3+}>\text{Mn}^{3+} \)

Hint:

In low spin octahedral complexes, electrons pair up in \(t_{2g}\) orbitals before occupying \(e_g\) orbitals.

Answer 1

The Correct Option is A

To determine the decreasing order of unpaired electrons in the low spin octahedral complexes of the transition metal ions \( \text{Mn}^{3+}, \text{Cr}^{3+}, \text{Fe}^{3+}, \) and \( \text{Co}^{3+} \), we need to consider their electronic configurations and the effect of crystal field splitting in a low spin configuration.

In a low spin octahedral complex, the strong field ligands cause a significant splitting of the \(d\)-orbitals, leading to a preference for electrons to pair up in the lower energy \(t_{2g}\) orbitals before occupying the higher energy \(e_g\) orbitals. Let's examine each ion:

1. \(\text{Cr}^{3+}\):
   • Electronic configuration of \(\text{Cr}\) is \([\text{Ar}] 3d^5 4s^1\).
   • For \(\text{Cr}^{3+}\), three electrons are removed, typically from \(4s\) and \(3d\), resulting in \(3d^3\).
   • In a low spin octahedral complex, all three electrons will be unpaired in the \(t_{2g}\) orbitals: \((t_{2g})^{3}\)
   • Number of unpaired electrons: \(3\).
2. \(\text{Mn}^{3+}\):
   • Electronic configuration of \(\text{Mn}\) is \([\text{Ar}] 3d^5 4s^2\).
   • For \(\text{Mn}^{3+}\), removing three electrons results in \(3d^4\).
   • In a low spin octahedral complex, electrons pair in the \(t_{2g}\) set: \((t_{2g})^{4}\).
   • Number of unpaired electrons: \(2\).
3. \(\text{Fe}^{3+}\):
   • Electronic configuration of \(\text{Fe}\) is \([\text{Ar}] 3d^6 4s^2\).
   • For \(\text{Fe}^{3+}\), removing three electrons results in \(3d^5\).
   • In a low spin octahedral complex, electrons pair in the \(t_{2g}\) set: \((t_{2g})^{5}\).
   • Number of unpaired electrons: \(1\).
4. \(\text{Co}^{3+}\):
   • Electronic configuration of \(\text{Co}\) is \([\text{Ar}] 3d^7 4s^2\).
   • For \(\text{Co}^{3+}\), removing three electrons results in \(3d^6\).
   • In a low spin octahedral complex, fully paired in \(t_{2g}\): \((t_{2g})^{6}\).
   • Number of unpaired electrons: \(0\).

Based on the calculated number of unpaired electrons:

Ion	Unpaired Electrons
\( \text{Cr}^{3+} \)	3
\( \text{Mn}^{3+} \)	2
\( \text{Fe}^{3+} \)	1
\( \text{Co}^{3+} \)	0

Thus, the correct decreasing order of unpaired electrons in their respective \( d \)-orbitals of the complexes is \(\text{Cr}^{3+} > \text{Mn}^{3+} > \text{Fe}^{3+} > \text{Co}^{3+}\). This matches the given correct answer.