Question

Find the ratio of CFSE of [Co(C$_2$O$_4$)$_3$]$^{3-}$ and [Cr(C$_2$O$_4$)$_3$]$^{3-}$, for each complex assume \(\Delta_t>P\):

• A. 6/3 = 2
• B. 6/5 = 1.2
• C. 3/5 = 0.6
• D. 1/3 = 0.33

Hint:

The CFSE depends on the distribution of electrons in the t$_{2g}$ and e$_g$ orbitals. For \( \Delta_t>P \), the metal ion configuration will determine the CFSE and allow for comparisons between different complexes.

Answer 1

The Correct Option is B

Step 1: Write the electronic configuration for Co$^{3+}$ and Cr$^{3+}$.
- Co$^{3+}$ has the electronic configuration 3d$^6$. The 3d orbitals split into t$_{2g}$ and e$_g$ orbitals, and the CFSE can be calculated as: \[ \text{CFSE (Co$^{3+}$)} = 6 \times \left( -\frac{2}{5} \right) \Delta_0 = -\frac{12}{5} \Delta_0 \] - Cr$^{3+}$ has the electronic configuration 3d$^3$. The CFSE for Cr$^{3+}$ is: \[ \text{CFSE (Cr$^{3+}$)} = 3 \times \left( -\frac{2}{5} \right) \Delta_0 = -\frac{6}{5} \Delta_0 \]
Step 2: Calculate the ratio of CFSE.
The ratio of CFSEs is: \[ \frac{\text{CFSE (Co$^{3+}$)}}{\text{CFSE (Cr$^{3+}$)}} = \frac{-\frac{12}{5} \Delta_0}{-\frac{6}{5} \Delta_0} = \frac{12}{6} = 2 \]
Step 3: Conclusion.
The ratio of CFSE is 2, so the correct answer is (2).