Question

Number of unpaired electrons in low spin octahedral complex formed by ions Mn$^{3+}$, Cr$^{3+}$, Fe$^{3+}$, and Co$^{3+}$ follows the order:

• A. Mn$^{3+}$>Cr$^{3+}$>Fe$^{3+}$>Co$^{3+}$
• B. Mn$^{3+}$>Fe$^{3+}$>Co$^{3+}$>Cr$^{3+}$
• C. Fe$^{3+}$>Cr$^{3+}$>Co$^{3+}$>Mn$^{3+}$
• D. Co$^{3+}$>Fe$^{3+}$>Cr$^{3+}$>Mn$^{3+}$

Hint:

In octahedral complexes, low spin configuration results in pairing of electrons in t$_{2g}$ and e$_g$ orbitals, influencing the number of unpaired electrons.

Answer 1

The Correct Option is B

Step 1: Electronic configuration of Co$^{3+$:}
Co$^{3+}$ → 3d$^6$ → t$_{2g}^6$ e$_g^0$, unpaired electron = 0

Step 2: Electronic configuration of Fe$^{3+$:}
Fe$^{3+}$ → 3d$^5$ → t$_{2g}^3$ e$_g^2$, unpaired electron = 1

Step 3: Electronic configuration of Cr$^{3+$:}
Cr$^{3+}$ → 3d$^3$ → t$_{2g}^3$ e$_g^0$, unpaired electron = 3

Step 4: Electronic configuration of Mn$^{3+$:}
Mn$^{3+}$ → 3d$^4$ → t$_{2g}^3$ e$_g^1$, unpaired electron = 2

Step 5: Conclusion.
The order of unpaired electrons is Mn$^{3+}$>Fe$^{3+}$>Co$^{3+}$>Cr$^{3+}$, which corresponds to option (2).