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

Step 1: Determine electronic configuration of each ion.
\[ \text{Cr}^{3+}: d^3,\quad \text{Mn}^{3+}: d^4,\quad \text{Fe}^{3+}: d^5,\quad \text{Co}^{3+}: d^6 \] All complexes are low spin octahedral.
Step 2: Count unpaired electrons in low spin case.
\[ \text{Cr}^{3+} (d^3): 3 \text{ unpaired} \] \[ \text{Mn}^{3+} (d^4): 2 \text{ unpaired} \] \[ \text{Fe}^{3+} (d^5): 1 \text{ unpaired} \] \[ \text{Co}^{3+} (d^6): 0 \text{ unpaired} \] Step 3: Arrange in decreasing order.
\[ \text{Cr}^{3+}>\text{Mn}^{3+}>\text{Fe}^{3+}>\text{Co}^{3+} \]
Final Answer: \[ \boxed{\text{Cr}^{3+}>\text{Mn}^{3+}>\text{Fe}^{3+}>\text{Co}^{3+}} \]