How is the tetrachloridocobaltate(II) anion stable if its crystal field stabilization energy comes to be positive?

I tried to calculate the ligand field stabilisation enthalpy of $\ce{[CoCl4]^2-}$ (as shown in the image) but found it to be positive. What could be the reason behind this? How is it stable with positive crystal field stabilization energy?


![enter image description here](https://i.stack.imgur.com/wAkwb.jpg)




You didnt put the right values. Higher orbitals have energy value of $+0.4 \Delta\_t$ and the lower orbitals have the energy value of $-0.6\Delta\_t$.


This complex has a coordination number of 4, so it can have tetrahedral or square planar geometry. Tetrahedral complexes are favored where attainment of regular complexes is important. For the tetrahedral complexes $\mathrm{d^0, d^2, d^5, d^7}$, and $\mathrm{d^{10}}$ the configurations are regular.


In tetrahedral complexes, mostly $\Delta t < P\ (\Delta\_t \approx 4/9 \Delta\_o)$ and hence they generally form high spin complexes.


[![Tetrahedral field split](https://i.stack.imgur.com/SBi8p.gif)](https://i.stack.imgur.com/SBi8p.gif)


The splitting will be into $3\ \mathrm{t\_2}$ (not $\mathrm{t\_{2g}}$) higher energy orbitals and $2\ \mathrm e$ (not $\mathrm{e\_g}$) lower energy orbitals. The effect of the ligand wont matter here because the value of $\Delta\_t$ is small. So the filling will be same for strong field and weak field. 


$$\begin{align}\text{CFSE} &= +0.4(3) -0.6(4)\\
&= -1.2\end{align}$$


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