The crystal field splitting energy of $\left[\mathrm{Co}(\text { oxalate })_{3}\right]^{3-}$ complex is ' $\mathrm{n}^{\prime}$ times that of the $\left[\mathrm{Cr}(\text { oxalate })_{3}\right]^{3-}$ complex. Here ' $\mathrm{n}^{\prime}$ is $\_\_\_\_$. (Assume $\Delta_{0} \gg \mathrm{P}$)

The Crystal Field Stabilization Energy (CFSE) is calculated based on the electronic configuration of the central metal ion in the octahedral field.
For $[Co(ox)_3]^{3-}$:
Cobalt is in $+3$ oxidation state. $Co^{3+}$ has a $d^6$ configuration.
Given $\Delta_0 \gg P$, it is a low spin complex. The electrons fill the $t_{2g}$ level first.
Configuration: $t_{2g}^6 e_g^0$.
CFSE for $Co^{3+} = 6 \times (-0.4 \Delta_0) + 0 \times (0.6 \Delta_0) = -2.4 \Delta_0$.
For $[Cr(ox)_3]^{3-}$:
Chromium is in $+3$ oxidation state. $Cr^{3+}$ has a $d^3$ configuration.
Configuration: $t_{2g}^3 e_g^0$.
CFSE for $Cr^{3+} = 3 \times (-0.4 \Delta_0) + 0 \times (0.6 \Delta_0) = -1.2 \Delta_0$.
Comparing the magnitudes of the stabilization energies:
Magnitude of CFSE for $Co^{3+} = 2.4 \Delta_0$.
Magnitude of CFSE for $Cr^{3+} = 1.2 \Delta_0$.
The ratio $n = \frac{2.4 \Delta_0}{1.2 \Delta_0} = 2$.